\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{The equation of a circle of radius }5\text{ units which touches the coordinate}
\displaystyle \text{axes in the second quadrant is:}
\displaystyle \text{(a) }x^2+y^2+10x-10y+25=0
\displaystyle \text{(b) }x^2+y^2-10x-10y+25=0
\displaystyle \text{(c) }x^2+y^2+10x+10y+25=0
\displaystyle \text{(d) }x^2+y^2-10x-10y-25=0
\displaystyle \text{Answer:}
\displaystyle \text{Since the circle touches both coordinate axes and lies in the second quadrant,}
\displaystyle \text{its centre is }(-5,5).
\displaystyle \text{Radius }=5.
\displaystyle \text{Using }(x-h)^2+(y-k)^2=r^2,
\displaystyle (x+5)^2+(y-5)^2=25.
\displaystyle x^2+10x+25+y^2-10y+25=25.
\displaystyle \therefore x^2+y^2+10x-10y+25=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The equation of the circle passing through the origin and cutting intercepts }2
\displaystyle \text{and }4\text{ on the coordinate axes is:}
\displaystyle \text{(a) }x^2+y^2-2x-4y=0
\displaystyle \text{(b) }x^2+y^2+2x+4y=0
\displaystyle \text{(c) }x^2+y^2+4x+8y=0
\displaystyle \text{(d) }x^2+y^2-4x-8y=0
\displaystyle \text{Answer:}
\displaystyle \text{The general equation of a circle through the origin is}
\displaystyle x^2+y^2+Dx+Ey=0.
\displaystyle \text{Since the circle cuts an intercept }2\text{ on the }x\text{-axis, it passes through }(2,0).
\displaystyle 2^2+2D=0.
\displaystyle 4+2D=0\Rightarrow D=-2.
\displaystyle \text{Since it cuts an intercept }4\text{ on the }y\text{-axis, it passes through }(0,4).
\displaystyle 4^2+4E=0.
\displaystyle 16+4E=0\Rightarrow E=-4.
\displaystyle \therefore x^2+y^2-2x-4y=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The equation }(4m-3)x^2+my^2+10x-8y+16=0\text{ represents a}
\displaystyle \text{circle for }m=:
\displaystyle \text{(a) }3\qquad\text{(b) }\frac{3}{4}\qquad\text{(c) }1\qquad\text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle \text{For the equation to represent a circle, the coefficients of }x^2\text{ and }y^2\text{ must be equal.}
\displaystyle \therefore 4m-3=m.
\displaystyle 3m=3.
\displaystyle \therefore m=1.
\displaystyle \text{For }m=1,\text{ the equation becomes}
\displaystyle x^2+y^2+10x-8y+16=0.
\displaystyle (x+5)^2+(y-4)^2=25.
\displaystyle \text{Hence, it represents a real circle of radius }5.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The radius of the circle }x^2+y^2+8x+10y-8=0\text{ is:}
\displaystyle \text{(a) }8\qquad\text{(b) }10\qquad\text{(c) }9\qquad\text{(d) }7
\displaystyle \text{Answer:}
\displaystyle x^2+y^2+8x+10y-8=0.
\displaystyle (x^2+8x)+(y^2+10y)=8.
\displaystyle (x+4)^2-16+(y+5)^2-25=8.
\displaystyle (x+4)^2+(y+5)^2=49.
\displaystyle \therefore r=\sqrt{49}=7.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The centre of the circle }x^2+y^2+8x+10y-8=0\text{ is:}
\displaystyle \text{(a) }(-4,-5)\qquad\text{(b) }(4,5)\qquad\text{(c) }(-8,-10)\qquad\text{(d) }(8,10)
\displaystyle \text{Answer:}
\displaystyle \text{The general equation of a circle is }x^2+y^2+2gx+2fy+c=0.
\displaystyle \text{Comparing, }2g=8\Rightarrow g=4.
\displaystyle 2f=10\Rightarrow f=5.
\displaystyle \therefore \text{Centre}=(-g,-f)=(-4,-5).
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the equation of the circle with centre }(-3,2)\text{ and radius }4.
\displaystyle \text{(a) }(x-3)^2+(y-2)^2=16
\displaystyle \text{(b) }(x+3)^2+(y+2)^2=16
\displaystyle \text{(c) }(x-3)^2+(y+2)^2=16
\displaystyle \text{(d) }(x+3)^2+(y-2)^2=16
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a circle with centre }(h,k)\text{ and radius }r\text{ is}
\displaystyle (x-h)^2+(y-k)^2=r^2.
\displaystyle \text{Here, }h=-3,\ k=2,\ r=4.
\displaystyle \therefore (x+3)^2+(y-2)^2=16.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The eccentricity of a circle is:}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }1.5\qquad\text{(d) }0.5
\displaystyle \text{Answer:}
\displaystyle \text{The eccentricity of a circle is }0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The eccentricity of a parabola is:}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }1.5\qquad\text{(d) }0.5
\displaystyle \text{Answer:}
\displaystyle \text{For a parabola, the distance of any point from the focus is equal to its}
\displaystyle \text{perpendicular distance from the directrix.}
\displaystyle \therefore \text{Eccentricity }e=1.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the equation of the parabola with vertex at }(0,0)\text{ and focus at}
\displaystyle (0,2).
\displaystyle \text{(a) }x^2=8y\qquad\text{(b) }x^2=-8y\qquad\text{(c) }y^2=8x\qquad\text{(d) }y^2=-8x
\displaystyle \text{Answer:}
\displaystyle \text{The standard equation of a parabola with vertex }(0,0)\text{ and focus }(0,a)\text{ is}
\displaystyle x^2=4ay.
\displaystyle \text{Here, the focus is }(0,2).
\displaystyle \therefore a=2.
\displaystyle x^2=4(2)y=8y.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The directrix of the parabola }y^2=48x\text{ is:}
\displaystyle \text{(a) }x=-12\qquad\text{(b) }x=12\qquad\text{(c) }x=24\qquad\text{(d) }x=-24
\displaystyle \text{Answer:}
\displaystyle \text{The standard equation of the parabola is }y^2=4ax.
\displaystyle \text{Comparing }y^2=48x\text{ with }y^2=4ax,
\displaystyle 4a=48.
\displaystyle \therefore a=12.
\displaystyle \text{The directrix of }y^2=4ax\text{ is }x=-a.
\displaystyle \therefore x=-12.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{What point of the parabola }x^2=9y\text{ has the abscissa three times that}
\displaystyle \text{of the ordinate?}
\displaystyle \text{(a) }(1,1)\qquad\text{(b) }(3,1)\qquad\text{(c) }(-3,-1)\qquad\text{(d) }(-3,-3)
\displaystyle \text{Answer:}
\displaystyle \text{Abscissa is three times the ordinate.}
\displaystyle \therefore x=3y.
\displaystyle \text{Substituting in }x^2=9y,
\displaystyle (3y)^2=9y.
\displaystyle 9y^2=9y.
\displaystyle y(y-1)=0.
\displaystyle y=0\text{ or }1.
\displaystyle y=0\text{ gives }(0,0),\text{ which is not among the options.}
\displaystyle y=1\Rightarrow x=3.
\displaystyle \therefore \text{The required point is }(3,1).
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The point where the axis meets the parabola is called:}
\displaystyle \text{(a) Latus rectum}\qquad\text{(b) Tangent}\qquad\text{(c) Chord}\qquad\text{(d) Vertex}
\displaystyle \text{Answer:}
\displaystyle \text{The axis of a parabola passes through its turning point.}
\displaystyle \text{This point is called the vertex.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The vertex of the parabola }(y+a)^2=8a(x-a)\text{ is:}
\displaystyle \text{(a) }(-a,-a)\qquad\text{(b) }(a,a)\qquad\text{(c) }(-a,a)\qquad\text{(d) }(a,-a)
\displaystyle \text{Answer:}
\displaystyle \text{The standard form of a parabola is }(y-k)^2=4A(x-h).
\displaystyle \text{Its vertex is }(h,k).
\displaystyle (y+a)^2=8a(x-a).
\displaystyle \therefore (y-(-a))^2=8a(x-a).
\displaystyle \therefore h=a\text{ and }k=-a.
\displaystyle \therefore \text{The vertex is }(a,-a).
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The length of the latus rectum of the parabola }y^2=16x\text{ is:}
\displaystyle \text{(a) }-16\qquad\text{(b) }-4\qquad\text{(c) }16\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \text{The standard equation of a parabola is }y^2=4ax.
\displaystyle \text{Comparing }y^2=16x\text{ with }y^2=4ax,
\displaystyle 4a=16\Rightarrow a=4.
\displaystyle \text{Length of the latus rectum}=4a.
\displaystyle =4(4)=16.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The focus of the parabola }x^2=16y\text{ is:}
\displaystyle \text{(a) }(-4,0)\qquad\text{(b) }(4,0)\qquad\text{(c) }(0,4)\qquad\text{(d) }(0,-4)
\displaystyle \text{Answer:}
\displaystyle \text{The standard equation of a parabola is }x^2=4ay.
\displaystyle \text{Comparing }x^2=16y\text{ with }x^2=4ay,
\displaystyle 4a=16\Rightarrow a=4.
\displaystyle \text{The focus of }x^2=4ay\text{ is }(0,a).
\displaystyle \therefore \text{The focus is }(0,4).
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In an ellipse, the distance between its foci is }6\text{ and its minor axis}
\displaystyle \text{is }8.\text{ Then its eccentricity is:}
\displaystyle \text{(a) }\frac{4}{5}\qquad\text{(b) }\frac{1}{\sqrt{52}}\qquad\text{(c) }\frac{3}{5}\qquad\text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Distance between the foci}=2c=6.
\displaystyle \therefore c=3.
\displaystyle \text{Length of the minor axis}=2b=8.
\displaystyle \therefore b=4.
\displaystyle \text{For an ellipse, }c^2=a^2-b^2.
\displaystyle \therefore a^2=b^2+c^2=4^2+3^2=25.
\displaystyle \therefore a=5.
\displaystyle \text{Eccentricity }e=\frac{c}{a}=\frac{3}{5}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The difference between the length of the major axis and latus rectum}
\displaystyle \text{of an ellipse is:}
\displaystyle \text{(a) }2ae^2\qquad\text{(b) }ae^2\qquad\text{(c) }2ae\qquad\text{(d) }ae
\displaystyle \text{Answer:}
\displaystyle \text{Length of the major axis}=2a.
\displaystyle \text{Length of the latus rectum}=\frac{2b^2}{a}.
\displaystyle \therefore \text{Required difference}=2a-\frac{2b^2}{a}.
\displaystyle =\frac{2(a^2-b^2)}{a}.
\displaystyle \text{For an ellipse, }a^2-b^2=c^2.
\displaystyle \therefore \text{Required difference}=\frac{2c^2}{a}.
\displaystyle \text{Since }e=\frac{c}{a},\text{ we have }c=ae.
\displaystyle \therefore \text{Required difference}=\frac{2a^2e^2}{a}=2ae^2.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The eccentricity of an ellipse, if the minor axis is equal to the distance}
\displaystyle \text{between the foci, is:}
\displaystyle \text{(a) }\frac{\sqrt3}{2}\qquad\text{(b) }\frac{2}{\sqrt3}\qquad\text{(c) }\frac{\sqrt2}{3}\qquad\text{(d) }\frac{1}{\sqrt2}
\displaystyle \text{Answer:}
\displaystyle \text{Length of the minor axis}=2b.
\displaystyle \text{Distance between the foci}=2c.
\displaystyle \text{According to the given condition,}
\displaystyle 2b=2c\Rightarrow b=c.
\displaystyle \text{For an ellipse, }c^2=a^2-b^2.
\displaystyle \text{Since }b=c,
\displaystyle c^2=a^2-c^2.
\displaystyle \therefore a^2=2c^2.
\displaystyle \therefore \frac{c^2}{a^2}=\frac{1}{2}.
\displaystyle \text{Since }e=\frac{c}{a},
\displaystyle e^2=\frac{1}{2}.
\displaystyle \therefore e=\frac{1}{\sqrt2}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The eccentricity of the ellipse }5x^2+9y^2=1\text{ is:}
\displaystyle \text{(a) }\frac{3}{4}\qquad\text{(b) }\frac{4}{5}\qquad\text{(c) }\frac{2}{3}\qquad\text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle 5x^2+9y^2=1.
\displaystyle \therefore \frac{x^2}{1/5}+\frac{y^2}{1/9}=1.
\displaystyle \text{Comparing with }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,
\displaystyle a^2=\frac{1}{5},\qquad b^2=\frac{1}{9}.
\displaystyle \text{For an ellipse, }e=\sqrt{1-\frac{b^2}{a^2}}.
\displaystyle \therefore e=\sqrt{1-\frac{1/9}{1/5}}.
\displaystyle =\sqrt{1-\frac{5}{9}}=\sqrt{\frac{4}{9}}=\frac{2}{3}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The eccentricity of a rectangular hyperbola is:}
\displaystyle \text{(a) }1\qquad\text{(b) }\sqrt2\qquad\text{(c) }\sqrt3\qquad\text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \text{For the hyperbola }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,
\displaystyle e=\sqrt{1+\frac{b^2}{a^2}}.
\displaystyle \text{For a rectangular hyperbola, }a=b.
\displaystyle \therefore e=\sqrt{1+\frac{a^2}{a^2}}.
\displaystyle =\sqrt2.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \text{CASE BASED/SOURCE BASED / PASSAGE BASED}

\displaystyle \textbf{Question 21: }\text{Consider the family of curves }C:x^2+y^2-2x-2ay-8=0,
\displaystyle \text{where }a\text{ is a variable. Gaurang is analysing the equation of the curve and its}
\displaystyle \text{dependence on }a.\text{ On the basis of the given equation, answer the following questions:}

\displaystyle \text{(i) The equation represents a family of circles having centre in the first quadrant if }a\text{ is:}
\displaystyle \text{(a) Any real number}\qquad\text{(b) }0
\displaystyle \text{(c) Always a negative real number}\qquad\text{(d) Always a positive real number}
\displaystyle \text{Answer:}
\displaystyle x^2+y^2-2x-2ay-8=0.
\displaystyle (x-1)^2+(y-a)^2=a^2+9.
\displaystyle \therefore \text{Centre of the circle is }(1,a).
\displaystyle \text{For the centre to lie in the first quadrant, }1>0\text{ and }a>0.
\displaystyle \therefore a\text{ must be positive.}
\displaystyle \therefore \text{Option (d) is correct.}

\displaystyle \text{(ii) If }a=0,\text{ then the centre of }C\text{ is:}
\displaystyle \text{(a) }(1,0)\qquad\text{(b) }(0,1)\qquad\text{(c) }(8,0)\qquad\text{(d) }(0,8)
\displaystyle \text{Answer:}
\displaystyle \text{Centre of the circle is }(1,a).
\displaystyle \text{For }a=0,\text{ centre}=(1,0).
\displaystyle \therefore \text{Option (a) is correct.}

\displaystyle \text{(iii) If }a=0,\text{ then the radius of }C\text{ is:}
\displaystyle \text{(a) }1\qquad\text{(b) }3\qquad\text{(c) }2\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \text{Radius}=\sqrt{a^2+9}.
\displaystyle \text{For }a=0,\quad r=\sqrt{9}=3.
\displaystyle \therefore \text{Option (b) is correct.}

\displaystyle \text{(iv) The curve }C\text{ passes through two fixed points whose coordinates are:}
\displaystyle \text{(a) }(-2,0),(4,0)\qquad\text{(b) }(2,0),(4,0)
\displaystyle \text{(c) }(-4,0),(4,0)\qquad\text{(d) }(2,0),(-4,0)
\displaystyle \text{Answer:}
\displaystyle x^2+y^2-2x-2ay-8=0.
\displaystyle \text{For fixed points independent of }a,\text{ the coefficient of }a\text{ must vanish.}
\displaystyle \therefore -2ay=0\Rightarrow y=0.
\displaystyle \text{Substituting }y=0,
\displaystyle x^2-2x-8=0.
\displaystyle (x-4)(x+2)=0.
\displaystyle \therefore x=4\text{ or }x=-2.
\displaystyle \therefore \text{The fixed points are }(-2,0)\text{ and }(4,0).
\displaystyle \therefore \text{Option (a) is correct.}

\displaystyle \text{(v) The eccentricity of the curve }C\text{ is:}
\displaystyle \text{(a) Depends on the value of }a\qquad\text{(b) Always }0
\displaystyle \text{(c) Always }1\qquad\text{(d) Always less than }1
\displaystyle \text{Answer:}
\displaystyle \text{For every real value of }a,\text{ the equation represents a circle.}
\displaystyle \text{The eccentricity of a circle is }0.
\displaystyle \therefore \text{The eccentricity of }C\text{ is always }0.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Case Study: The focus of a parabolic mirror, as shown in the figure, is at a}
\displaystyle \text{distance of }5\text{ cm from its vertex and the mirror is }45\text{ cm deep. Based on the above}
\displaystyle \text{information, answer the following questions:}

\displaystyle \text{(I) What is the equation of the parabolic section?}
\displaystyle \text{(a) }y^2=-5x\qquad\text{(b) }y^2=45x\qquad\text{(c) }y^2=5x\qquad\text{(d) }y^2=20x
\displaystyle \text{Answer:}
\displaystyle \text{The vertex is }O(0,0)\text{ and the focus is }(5,0).
\displaystyle \therefore a=5.
\displaystyle \text{The standard equation of the parabola is }y^2=4ax.
\displaystyle \therefore y^2=4(5)x=20x.
\displaystyle \therefore \text{Option (d) is correct.}

\displaystyle \text{(II) What is the length of }AB\text{?}
\displaystyle \text{(a) }5\text{ cm}\qquad\text{(b) }45\text{ cm}\qquad\text{(c) }60\text{ cm}\qquad\text{(d) }20\text{ cm}
\displaystyle \text{Answer:}
\displaystyle \text{The mirror is }45\text{ cm deep, so the rim }AB\text{ lies on }x=45.
\displaystyle \text{Using }y^2=20x,
\displaystyle y^2=20(45)=900.
\displaystyle \therefore y=\pm30.
\displaystyle \therefore A=(45,30)\text{ and }B=(45,-30).
\displaystyle AB=30-(-30)=60\text{ cm}.
\displaystyle \therefore \text{The correct length of }AB\text{ is }60\text{ cm.}
\displaystyle \therefore \text{Option (c) is correct.}

\displaystyle \text{(III) What are the coordinates of the focus of the parabolic mirror?}
\displaystyle \text{(a) }(5,0)\qquad\text{(b) }(20,0)\qquad\text{(c) }(45,30)\qquad\text{(d) }(45,15)
\displaystyle \text{Answer:}
\displaystyle \text{The focus is }5\text{ cm from the vertex along the positive }x\text{-axis.}
\displaystyle \therefore \text{The focus is }(5,0).
\displaystyle \therefore \text{Option (a) is correct.}

\displaystyle \text{(IV) What are the coordinates of point }A\text{ of the parabolic mirror?}
\displaystyle \text{(a) }(5,0)\qquad\text{(b) }(45,5)\qquad\text{(c) }(45,30)\qquad\text{(d) }(45,15)
\displaystyle \text{Answer:}
\displaystyle \text{At the rim of the mirror, }x=45.
\displaystyle y^2=20(45)=900.
\displaystyle \therefore y=\pm30.
\displaystyle \text{Since }A\text{ is the upper point, }y=30.
\displaystyle \therefore A=(45,30).
\displaystyle \therefore \text{Option (c) is correct.}

\displaystyle \text{(V) What is the equation of the axis of the parabolic mirror?}
\displaystyle \text{(a) }x=0\qquad\text{(b) }y=0\qquad\text{(c) }x=45\qquad\text{(d) }y=45
\displaystyle \text{Answer:}
\displaystyle \text{The parabola }y^2=20x\text{ has the }x\text{-axis as its axis of symmetry.}
\displaystyle \therefore \text{The equation of the axis is }y=0.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{The standard form of the equation of an ellipse with centre }(h,k)\text{ and}
\displaystyle \text{major axis parallel to the }x\text{-axis is}
\displaystyle \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1,\qquad a>b.
\displaystyle \text{The length of the major axis is }2a\text{ and the coordinates of the vertices are } \\ (h\pm a,k).
\displaystyle \text{The length of the minor axis is }2b\text{ and the coordinates of the co-vertices are } \\ (h,k\pm b).
\displaystyle \text{The coordinates of the foci are }(h\pm c,k),\text{ where }c^2=a^2-b^2.
\displaystyle \text{If the major axis is parallel to the }y\text{-axis, the standard form is}
\displaystyle \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1,\qquad a>b.
\displaystyle \text{The vertices are }(h,k\pm a),\text{ the co-vertices are }(h\pm b,k),\text{ and the foci are}
\displaystyle (h,k\pm c),\text{ where }c^2=a^2-b^2.
\displaystyle \text{Based on the above information, answer the following questions:}

\displaystyle \text{(I) If the equation of the ellipse is }(x-2)^2+9(y-1)^2=9,\text{ then the length of its}
\displaystyle \text{major axis is:}
\displaystyle \text{(a) }3\qquad\text{(b) }6\qquad\text{(c) }9\qquad\text{(d) }18
\displaystyle \text{Answer:}
\displaystyle (x-2)^2+9(y-1)^2=9.
\displaystyle \therefore \frac{(x-2)^2}{9}+\frac{(y-1)^2}{1}=1.
\displaystyle \therefore a^2=9\Rightarrow a=3.
\displaystyle \text{Length of the major axis}=2a=6.
\displaystyle \therefore \text{Option (b) is correct.}

\displaystyle \text{(II) If the equation of the ellipse is }(x-2)^2+9(y-1)^2=9,\text{ then its centre is:}
\displaystyle \text{(a) }(2,1)\qquad\text{(b) }(3,1)\qquad\text{(c) }(-2,1)\qquad\text{(d) }(1,9)
\displaystyle \text{Answer:}
\displaystyle \frac{(x-2)^2}{9}+\frac{(y-1)^2}{1}=1.
\displaystyle \text{Comparing with the standard form, }h=2\text{ and }k=1.
\displaystyle \therefore \text{The centre is }(2,1).
\displaystyle \therefore \text{Option (a) is correct.}

\displaystyle \text{(III) If the equation of the ellipse is }9(x-2)^2+(y-1)^2=9,\text{ then its major}
\displaystyle \text{axis is parallel to the line:}
\displaystyle \text{(a) }x+y=0\qquad\text{(b) }y=0\qquad\text{(c) }x+y=0\qquad\text{(d) }x=0
\displaystyle \text{Answer:}
\displaystyle 9(x-2)^2+(y-1)^2=9.
\displaystyle \therefore \frac{(x-2)^2}{1}+\frac{(y-1)^2}{9}=1.
\displaystyle \text{The larger denominator is under the }y\text{-term.}
\displaystyle \therefore \text{The major axis is parallel to the }y\text{-axis.}
\displaystyle \text{The equation of the }y\text{-axis is }x=0.
\displaystyle \therefore \text{Option (d) is correct.}

\displaystyle \text{(IV) If the equation of the ellipse is }(x-2)^2+25(y-1)^2=25,\text{ then the coordinates}
\displaystyle \text{of its vertices are:}
\displaystyle \text{(a) }(2\pm5,1)\qquad\text{(b) }(2,1\pm1)
\displaystyle \text{(c) }(2\pm5,0)\qquad\text{(d) }(\pm5,0)
\displaystyle \text{Answer:}
\displaystyle (x-2)^2+25(y-1)^2=25.
\displaystyle \therefore \frac{(x-2)^2}{25}+\frac{(y-1)^2}{1}=1.
\displaystyle \therefore h=2,\quad k=1,\quad a=5.
\displaystyle \text{Since the major axis is parallel to the }x\text{-axis, the vertices are }(h\pm a,k).
\displaystyle \therefore \text{The vertices are }(2\pm5,1).
\displaystyle \therefore \text{Option (a) is correct.}

\displaystyle \text{(V) If the equation of the ellipse is }(x-2)^2+9(y-1)^2=9,\text{ then its foci are:}
\displaystyle \text{(a) }(2\pm2\sqrt2,1)\qquad\text{(b) }(2,1\pm2\sqrt2)
\displaystyle \text{(c) }(2,1)\qquad\text{(d) }(\pm2\sqrt2,0)
\displaystyle \text{Answer:}
\displaystyle \frac{(x-2)^2}{9}+\frac{(y-1)^2}{1}=1.
\displaystyle \therefore a^2=9,\qquad b^2=1.
\displaystyle c^2=a^2-b^2=9-1=8.
\displaystyle \therefore c=2\sqrt2.
\displaystyle \text{Since the major axis is parallel to the }x\text{-axis, the foci are }(h\pm c,k).
\displaystyle \therefore \text{The foci are }(2\pm2\sqrt2,1).
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \text{ASSERTION REASONING}

\displaystyle \textbf{Question 26: }\text{Assertion (A): The conic section having the equation }
\displaystyle 2x^2-2y^2-5=0 \ \text{is a circle.}
\displaystyle \text{Reason (R): The conic section having the general equation }
\displaystyle Ax^2+Cy^2+Dx+Ey+F=0 \  \text{is a circle if }A=C.

\displaystyle \text{Select the correct option from the following:}
\displaystyle \text{(a) Both A and R are true and R is the correct explanation of A.}
\displaystyle \text{(b) Both A and R are true but R is not the correct explanation of A.}
\displaystyle \text{(c) A is true but R is false.}
\displaystyle \text{(d) A is false but R is true.}
\displaystyle \text{Answer:}
\displaystyle 2x^2-2y^2-5=0.
\displaystyle \text{The coefficients of }x^2\text{ and }y^2\text{ are }2\text{ and }-2\text{ respectively.}
\displaystyle \text{Since they have opposite signs, the equation represents a hyperbola, not a circle.}
\displaystyle \therefore \text{Assertion (A) is false.}
\displaystyle \text{For }Ax^2+Cy^2+Dx+Ey+F=0\text{ to represent a circle, the coefficients of}
\displaystyle x^2\text{ and }y^2\text{ must be equal, i.e. }A=C.
\displaystyle \therefore \text{Reason (R) is true.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Assertion (A): The point }(1.5,2)\text{ is inside the circle}
\displaystyle S:x^2+y^2-5x-4y=0.
\displaystyle \text{Reason (R): The point }(x_1,y_1)\text{ is inside, on, or outside the circle}
\displaystyle S=x^2+y^2+2gx+2fy+c=0,\text{ according as }S_1<0,\ S_1=0,\text{ or }S_1>0,
\displaystyle \text{where }S_1=x_1^2+y_1^2+2gx_1+2fy_1+c.

\displaystyle \text{Select the correct option from the following:}
\displaystyle \text{(a) Both A and R are true and R is the correct explanation of A.}
\displaystyle \text{(b) Both A and R are true but R is not the correct explanation of A.}
\displaystyle \text{(c) A is true but R is false.}
\displaystyle \text{(d) A is false but R is true.}
\displaystyle \text{Answer:}
\displaystyle S=x^2+y^2-5x-4y.
\displaystyle \text{For the point }(1.5,2),
\displaystyle S_1=(1.5)^2+2^2-5(1.5)-4(2).
\displaystyle =\frac{9}{4}+4-\frac{15}{2}-8.
\displaystyle =-\frac{37}{4}<0.
\displaystyle \therefore (1.5,2)\text{ lies inside the circle.}
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{For a circle }S=0,\text{ a point lies inside, on, or outside the circle according as}
\displaystyle S_1<0,\quad S_1=0,\quad\text{or}\quad S_1>0,\text{ respectively.}
\displaystyle \therefore \text{Reason (R) is true and correctly explains Assertion (A).}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Assertion (A): The parabola }x^2=6y\text{ opens upwards along the}
\displaystyle y\text{-axis.}
\displaystyle \text{Reason (R): When the axis of symmetry is along the }x\text{-axis, the parabola opens upwards.}

\displaystyle \text{Select the correct option from the following:}
\displaystyle \text{(a) Both A and R are true and R is the correct explanation of A.}
\displaystyle \text{(b) Both A and R are true but R is not the correct explanation of A.}
\displaystyle \text{(c) A is true but R is false.}
\displaystyle \text{(d) A is false but R is true.}
\displaystyle \text{Answer:}
\displaystyle x^2=6y.
\displaystyle \text{Comparing with the standard form }x^2=4ay,
\displaystyle 4a=6\Rightarrow a=\frac{3}{2}>0.
\displaystyle \therefore \text{The parabola opens upwards along the positive }y\text{-axis.}
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{If the axis of symmetry is along the }x\text{-axis, the parabola opens either to the right}
\displaystyle \text{or to the left, depending on the sign in its equation.}
\displaystyle \therefore \text{Reason (R) is false.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Assertion (A): The major axis of the ellipse }5x^2+9y^2=1
\displaystyle \text{is along the }x\text{-axis.}
\displaystyle \text{Reason (R): The major axis is along the }x\text{-axis if }x^2\text{ has the larger denominator}
\displaystyle \text{and along the }y\text{-axis if }y^2\text{ has the larger denominator in the standard equation}
\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.

\displaystyle \text{Select the correct option from the following:}
\displaystyle \text{(a) Both A and R are true and R is the correct explanation of A.}
\displaystyle \text{(b) Both A and R are true but R is not the correct explanation of A.}
\displaystyle \text{(c) A is true but R is false.}
\displaystyle \text{(d) A is false but R is true.}
\displaystyle \text{Answer:}
\displaystyle 5x^2+9y^2=1.
\displaystyle \therefore \frac{x^2}{1/5}+\frac{y^2}{1/9}=1.
\displaystyle \text{Since }\frac{1}{5}>\frac{1}{9},\text{ the larger denominator is under }x^2.
\displaystyle \therefore \text{The major axis is along the }x\text{-axis.}
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{In the standard equation of an ellipse, the major axis corresponds to the variable}
\displaystyle \text{having the larger denominator.}
\displaystyle \therefore \text{Reason (R) is true and correctly explains Assertion (A).}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Assertion (A): The eccentricity of the hyperbola }
\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \ \text{is always greater than }1.
\displaystyle \text{Reason (R): For a hyperbola, }c^2=a^2+b^2\text{ and eccentricity }e=\frac{c}{a},
\displaystyle \text{so the eccentricity is always greater than }1.

\displaystyle \text{Select the correct option from the following:}
\displaystyle \text{(a) Both A and R are true and R is the correct explanation of A.}
\displaystyle \text{(b) Both A and R are true but R is not the correct explanation of A.}
\displaystyle \text{(c) A is true but R is false.}
\displaystyle \text{(d) A is false but R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{For the hyperbola }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,
\displaystyle c^2=a^2+b^2.
\displaystyle \therefore c^2>a^2.
\displaystyle \therefore c>a.
\displaystyle \text{Since }e=\frac{c}{a},
\displaystyle e>1.
\displaystyle \therefore \text{The eccentricity of a hyperbola is always greater than }1.
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{Reason (R) is also true and correctly explains Assertion (A).}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 31: }\text{Find the equation of the circle with centre }(-a,-a)\text{ and radius }\sqrt{2}\,a.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a circle with centre }(h,k)\text{ and radius }r\text{ is}
\displaystyle (x-h)^2+(y-k)^2=r^2.
\displaystyle \therefore (x+a)^2+(y+a)^2=(\sqrt{2}\,a)^2.
\displaystyle x^2+y^2+2ax+2ay+2a^2=2a^2.
\displaystyle \therefore x^2+y^2+2ax+2ay=0.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Find the equation of the circle whose endpoints of a diameter are }(a,0)
\displaystyle \text{and }(0,b).
\displaystyle \text{Answer:}
\displaystyle \text{If }(x_1,y_1)\text{ and }(x_2,y_2)\text{ are the endpoints of a diameter, then}
\displaystyle (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.
\displaystyle \therefore (x-a)(x-0)+(y-0)(y-b)=0.
\displaystyle \therefore x^2+y^2-ax-by=0.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Find the radius and centre of the circle } \\ 3x^2+3y^2+9x-6y+1=0.
\displaystyle \text{Answer:}
\displaystyle \text{Dividing by }3,
\displaystyle x^2+y^2+3x-2y+\frac{1}{3}=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,
\displaystyle g=\frac{3}{2},\qquad f=-1,\qquad c=\frac{1}{3}.
\displaystyle \therefore \text{Centre}=(-g,-f)=\left(-\frac{3}{2},1\right).
\displaystyle \text{Radius}=\sqrt{g^2+f^2-c}.
\displaystyle =\sqrt{\frac{9}{4}+1-\frac{1}{3}}=\sqrt{\frac{35}{12}}=\frac{\sqrt{105}}{6}.
\displaystyle \therefore \text{Centre}=\left(-\frac{3}{2},1\right),\qquad\text{radius}=\frac{\sqrt{105}}{6}.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Find the equation of the circle with centre }(1,5)\text{ and passing through}
\displaystyle \text{the point }(7,-1).
\displaystyle \text{Answer:}
\displaystyle r^2=(7-1)^2+(-1-5)^2.
\displaystyle =6^2+(-6)^2=72.
\displaystyle \therefore (x-1)^2+(y-5)^2=72.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Does the point }(7,-5)\text{ lie inside, outside or on the circle}
\displaystyle x^2+y^2-6x+8y+9=0\text{?}
\displaystyle \text{Answer:}
\displaystyle S_1=7^2+(-5)^2-6(7)+8(-5)+9.
\displaystyle =49+25-42-40+9=1.
\displaystyle \text{Since }S_1>0,\text{ the point lies outside the circle.}
\displaystyle \therefore (7,-5)\text{ lies outside the circle.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{For what value of }k\text{ does the equation }
\displaystyle 9x^2+y^2=k(x^2-y^2-2x) \ \text{represent the equation of a circle?}
\displaystyle \text{Answer:}
\displaystyle 9x^2+y^2=kx^2-ky^2-2kx.
\displaystyle (9-k)x^2+(1+k)y^2+2kx=0.
\displaystyle \text{For a circle, the coefficients of }x^2\text{ and }y^2\text{ must be equal.}
\displaystyle 9-k=1+k.
\displaystyle 2k=8.
\displaystyle \therefore k=4.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{Find the equation of the parabola with vertex at the origin and directrix}
\displaystyle y-2=0.
\displaystyle \text{Answer:}
\displaystyle \text{The directrix is }y=2.
\displaystyle \therefore \text{The focus is }(0,-2).
\displaystyle \text{Comparing with }x^2=-4ay,\text{ we have }a=2.
\displaystyle \therefore x^2=-8y.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Find the equation of the parabola with vertex at the origin, axis along the}
\displaystyle x\text{-axis and passing through the point }(3,2).
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the parabola is }y^2=4ax.
\displaystyle \text{Since it passes through }(3,2),
\displaystyle 2^2=4a(3).
\displaystyle 4=12a\Rightarrow a=\frac{1}{3}.
\displaystyle \therefore y^2=\frac{4}{3}x.
\displaystyle \therefore 3y^2=4x.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Find the equation of the parabola with focus }(3,-4)\text{ and directrix}
\displaystyle x+1=0.
\displaystyle \text{Answer:}
\displaystyle \text{The directrix is }x=-1\text{ and the focus is }(3,-4).
\displaystyle \text{The vertex is the midpoint between the focus and the directrix along the axis.}
\displaystyle \therefore \text{Vertex}=\left(\frac{3+(-1)}{2},-4\right)=(1,-4).
\displaystyle \text{Distance from the vertex to the focus}=3-1=2.
\displaystyle \therefore a=2.
\displaystyle \text{Using }(y-k)^2=4a(x-h),
\displaystyle (y+4)^2=8(x-1).
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Find the focus, vertex and directrix of the parabola }y^2=x.
\displaystyle \text{Answer:}
\displaystyle \text{The standard equation of a parabola is }y^2=4ax.
\displaystyle \text{Comparing }y^2=x\text{ with }y^2=4ax,
\displaystyle 4a=1\Rightarrow a=\frac{1}{4}.
\displaystyle \therefore \text{Focus}=\left(\frac{1}{4},0\right).
\displaystyle \therefore \text{Vertex}=(0,0).
\displaystyle \therefore \text{Directrix is }x=-\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{Determine the equation of the ellipse whose foci are }(4,0)\text{ and }(-4,0),
\displaystyle \text{and eccentricity is }\frac{1}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{The foci are }(\pm4,0),\text{ so }c=4.
\displaystyle \text{For an ellipse, }e=\frac{c}{a}.
\displaystyle \frac{1}{3}=\frac{4}{a}.
\displaystyle \therefore a=12\Rightarrow a^2=144.
\displaystyle \text{Also, }c^2=a^2-b^2.
\displaystyle 16=144-b^2.
\displaystyle \therefore b^2=128.
\displaystyle \text{Since the foci lie on the }x\text{-axis, the major axis is along the }x\text{-axis.}
\displaystyle \therefore \frac{x^2}{144}+\frac{y^2}{128}=1.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{If the distance between the foci of a hyperbola is }16\text{ and its eccentricity}
\displaystyle \text{is }\sqrt{2},\text{ obtain its equation.}
\displaystyle \text{Answer:}
\displaystyle \text{Distance between the foci}=2c=16.
\displaystyle \therefore c=8.
\displaystyle \text{For a hyperbola, }e=\frac{c}{a}.
\displaystyle \sqrt{2}=\frac{8}{a}.
\displaystyle \therefore a=\frac{8}{\sqrt{2}}=4\sqrt{2}.
\displaystyle \therefore a^2=32.
\displaystyle \text{For a hyperbola, }c^2=a^2+b^2.
\displaystyle 64=32+b^2.
\displaystyle \therefore b^2=32.
\displaystyle \therefore \frac{x^2}{32}-\frac{y^2}{32}=1.
\displaystyle \therefore x^2-y^2=32.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{In an ellipse, the distance between its foci is }10\text{ and its minor axis is}
\displaystyle \text{8. Find its eccentricity.}
\displaystyle \text{Answer:}
\displaystyle \text{Distance between the foci}=2c=10.
\displaystyle \therefore c=5.
\displaystyle \text{Length of the minor axis}=2b=8.
\displaystyle \therefore b=4.
\displaystyle \text{For an ellipse, }c^2=a^2-b^2.
\displaystyle \therefore a^2=b^2+c^2=4^2+5^2=41.
\displaystyle \therefore a=\sqrt{41}.
\displaystyle \text{Eccentricity }e=\frac{c}{a}=\frac{5}{\sqrt{41}}.
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{Find the equation of a diameter of the circle }
\displaystyle x^2+y^2+2x-4y=4 \ \text{which is parallel to the line }3x+5y=4.
\displaystyle \text{Answer:}
\displaystyle x^2+y^2+2x-4y-4=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,
\displaystyle g=1,\qquad f=-2.
\displaystyle \therefore \text{Centre}=(-g,-f)=(-1,2).
\displaystyle \text{The required diameter is parallel to }3x+5y=4.
\displaystyle \therefore \text{Its equation is of the form }3x+5y+c=0.
\displaystyle \text{Since it passes through }(-1,2),
\displaystyle 3(-1)+5(2)+c=0.
\displaystyle -3+10+c=0.
\displaystyle \therefore c=-7.
\displaystyle \therefore 3x+5y-7=0.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{Find the equation of a parabola whose vertex is at the origin, which is}
\displaystyle \text{symmetric about the }x\text{-axis and passes through the point }(-2,-3).
\displaystyle \text{Answer:}
\displaystyle \text{Since the parabola is symmetric about the }x\text{-axis and has vertex at the origin,}
\displaystyle \text{its equation is }y^2=4ax.
\displaystyle \text{Since it passes through }(-2,-3),
\displaystyle (-3)^2=4a(-2).
\displaystyle 9=-8a.
\displaystyle \therefore a=-\frac{9}{8}.
\displaystyle \therefore y^2=4\left(-\frac{9}{8}\right)x.
\displaystyle y^2=-\frac{9}{2}x.
\displaystyle \therefore 2y^2+9x=0.
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{If the eccentricity of an ellipse is }\frac{4}{9}\text{ and the distance between its}
\displaystyle \text{foci is }8\text{ units, find the length of the latus rectum of the ellipse.}
\displaystyle \text{Answer:}
\displaystyle \text{Distance between the foci}=2c=8.
\displaystyle \therefore c=4.
\displaystyle \text{For an ellipse, }e=\frac{c}{a}.
\displaystyle \frac{4}{9}=\frac{4}{a}.
\displaystyle \therefore a=9.
\displaystyle \text{Also, }c^2=a^2-b^2.
\displaystyle 16=81-b^2.
\displaystyle \therefore b^2=65.
\displaystyle \text{Length of the latus rectum}=\frac{2b^2}{a}.
\displaystyle =\frac{2(65)}{9}=\frac{130}{9}\text{ units}.
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{Find the foci of the hyperbola }9x^2-16y^2=144.
\displaystyle \text{Answer:}
\displaystyle 9x^2-16y^2=144.
\displaystyle \therefore \frac{x^2}{16}-\frac{y^2}{9}=1.
\displaystyle \therefore a^2=16,\qquad b^2=9.
\displaystyle \text{For a hyperbola, }c^2=a^2+b^2.
\displaystyle c^2=16+9=25.
\displaystyle \therefore c=5.
\displaystyle \therefore \text{The foci are }(\pm5,0).
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{Find the eccentricity of the hyperbola }3x^2-2y^2=1.
\displaystyle \text{Answer:}
\displaystyle 3x^2-2y^2=1.
\displaystyle \therefore \frac{x^2}{1/3}-\frac{y^2}{1/2}=1.
\displaystyle \therefore a^2=\frac{1}{3},\qquad b^2=\frac{1}{2}.
\displaystyle \text{For a hyperbola, }e=\sqrt{1+\frac{b^2}{a^2}}.
\displaystyle \therefore e=\sqrt{1+\frac{1/2}{1/3}}.
\displaystyle =\sqrt{1+\frac{3}{2}}=\sqrt{\frac{5}{2}}=\frac{\sqrt{10}}{2}.
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Find the equation of the hyperbola whose vertices are }(\pm3,0)\text{ and}
\displaystyle \text{foci are }(\pm5,0).
\displaystyle \text{Answer:}
\displaystyle \text{Since the vertices are }(\pm a,0),\text{ we have }a=3.
\displaystyle \therefore a^2=9.
\displaystyle \text{Since the foci are }(\pm c,0),\text{ we have }c=5.
\displaystyle \therefore c^2=25.
\displaystyle \text{For a hyperbola, }c^2=a^2+b^2.
\displaystyle 25=9+b^2.
\displaystyle \therefore b^2=16.
\displaystyle \therefore \text{The equation of the hyperbola is}
\displaystyle \frac{x^2}{9}-\frac{y^2}{16}=1.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{Find the length of the latus rectum of the hyperbola } \\ 16x^2-9y^2=144.
\displaystyle \text{Answer:}
\displaystyle 16x^2-9y^2=144.
\displaystyle \therefore \frac{x^2}{9}-\frac{y^2}{16}=1.
\displaystyle \therefore a^2=9,\qquad b^2=16.
\displaystyle \therefore a=3.
\displaystyle \text{Length of the latus rectum of a hyperbola}=\frac{2b^2}{a}.
\displaystyle =\frac{2(16)}{3}=\frac{32}{3}.
\displaystyle \therefore \text{The length of the latus rectum is }\frac{32}{3}\text{ units.}
\displaystyle \\

\displaystyle \text{SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 51: }\text{Find the length of the latus rectum of the parabola}
\displaystyle x^2-4x-8y+12=0.
\displaystyle \text{Answer:}
\displaystyle x^2-4x-8y+12=0.
\displaystyle x^2-4x+4=8y-8.
\displaystyle (x-2)^2=8(y-1).
\displaystyle \text{Comparing with }(x-h)^2=4a(y-k),
\displaystyle 4a=8\Rightarrow a=2.
\displaystyle \text{Length of the latus rectum}=4a=8.
\displaystyle \therefore \text{The length of the latus rectum is }8\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{Find the lengths of the axes, foci, vertices and eccentricity of the}
\displaystyle \text{hyperbola }2x^2-3y^2=6.
\displaystyle \text{Answer:}
\displaystyle 2x^2-3y^2=6.
\displaystyle \therefore \frac{x^2}{3}-\frac{y^2}{2}=1.
\displaystyle \text{Comparing with }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,
\displaystyle a^2=3,\qquad b^2=2.
\displaystyle \therefore a=\sqrt3,\qquad b=\sqrt2.
\displaystyle \text{Length of the transverse axis}=2a=2\sqrt3.
\displaystyle \text{Length of the conjugate axis}=2b=2\sqrt2.
\displaystyle \text{For a hyperbola, }c^2=a^2+b^2.
\displaystyle c^2=3+2=5\Rightarrow c=\sqrt5.
\displaystyle \therefore \text{The foci are }(\pm\sqrt5,0).
\displaystyle \therefore \text{The vertices are }(\pm\sqrt3,0).
\displaystyle \text{Eccentricity }e=\frac{c}{a}=\frac{\sqrt5}{\sqrt3}=\sqrt{\frac{5}{3}}.
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{Find the equation of the hyperbola whose foci are }(\pm3\sqrt5,0)\text{ and}
\displaystyle \text{length of the latus rectum is }8.
\displaystyle \text{Answer:}
\displaystyle \text{Since the foci are }(\pm3\sqrt5,0),
\displaystyle c=3\sqrt5\Rightarrow c^2=45.
\displaystyle \text{For the hyperbola }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,
\displaystyle c^2=a^2+b^2.
\displaystyle \therefore a^2+b^2=45.
\displaystyle \text{Length of the latus rectum}=\frac{2b^2}{a}=8.
\displaystyle \therefore b^2=4a.
\displaystyle a^2+4a=45.
\displaystyle a^2+4a-45=0.
\displaystyle (a-5)(a+9)=0.
\displaystyle \text{Since }a>0,\quad a=5.
\displaystyle \therefore a^2=25,\qquad b^2=20.
\displaystyle \therefore \frac{x^2}{25}-\frac{y^2}{20}=1.
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{For the ellipse }3x^2+2y^2=6,\text{ find the lengths of the major and minor}
\displaystyle \text{axes, eccentricity, foci, vertices and length of the latus rectum.}
\displaystyle \text{Answer:}
\displaystyle 3x^2+2y^2=6.
\displaystyle \therefore \frac{x^2}{2}+\frac{y^2}{3}=1.
\displaystyle \text{Since }3>2,\text{ the major axis is along the }y\text{-axis.}
\displaystyle \therefore a^2=3,\qquad b^2=2.
\displaystyle \therefore a=\sqrt3,\qquad b=\sqrt2.
\displaystyle \text{Length of the major axis}=2a=2\sqrt3.
\displaystyle \text{Length of the minor axis}=2b=2\sqrt2.
\displaystyle \text{For an ellipse, }c^2=a^2-b^2=3-2=1.
\displaystyle \therefore c=1.
\displaystyle \text{Eccentricity }e=\frac{c}{a}=\frac{1}{\sqrt3}.
\displaystyle \therefore \text{The foci are }(0,\pm1).
\displaystyle \therefore \text{The vertices are }(0,\pm\sqrt3).
\displaystyle \text{Length of the latus rectum}=\frac{2b^2}{a}.
\displaystyle =\frac{4}{\sqrt3}=\frac{4\sqrt3}{3}.
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{If the line }y=mx+1\text{ is tangent to the parabola }y^2=4x,\text{ find the}
\displaystyle \text{value of }m.
\displaystyle \text{Answer:}
\displaystyle y=mx+1.
\displaystyle \text{Substituting in }y^2=4x,
\displaystyle (mx+1)^2=4x.
\displaystyle m^2x^2+(2m-4)x+1=0.
\displaystyle \text{For the line to be tangent, the quadratic must have equal roots.}
\displaystyle \therefore (2m-4)^2-4m^2=0.
\displaystyle 4(m-2)^2-4m^2=0.
\displaystyle (m-2)^2=m^2.
\displaystyle m^2-4m+4=m^2.
\displaystyle \therefore m=1.
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{Find the equation of an ellipse whose eccentricity is }\frac{2}{3},\text{ length of}
\displaystyle \text{the latus rectum is }5\text{ and centre is }(0,0).
\displaystyle \text{Answer:}
\displaystyle \text{Given, }e=\frac{2}{3}.
\displaystyle \text{For an ellipse, }e^2=1-\frac{b^2}{a^2}.
\displaystyle \frac{4}{9}=1-\frac{b^2}{a^2}.
\displaystyle \therefore \frac{b^2}{a^2}=\frac{5}{9}.
\displaystyle \therefore b^2=\frac{5a^2}{9}.
\displaystyle \text{Length of the latus rectum}=\frac{2b^2}{a}=5.
\displaystyle \frac{2}{a}\left(\frac{5a^2}{9}\right)=5.
\displaystyle \frac{10a}{9}=5.
\displaystyle \therefore a=\frac{9}{2}.
\displaystyle \therefore a^2=\frac{81}{4}.
\displaystyle b^2=\frac{5}{9}\times\frac{81}{4}=\frac{45}{4}.
\displaystyle \text{If the major axis is along the }x\text{-axis,}
\displaystyle \frac{x^2}{81/4}+\frac{y^2}{45/4}=1.
\displaystyle \therefore 20x^2+36y^2=405.
\displaystyle \text{If the major axis is along the }y\text{-axis,}
\displaystyle \frac{x^2}{45/4}+\frac{y^2}{81/4}=1.
\displaystyle \therefore 36x^2+20y^2=405.
\displaystyle \therefore \text{Both equations are possible unless the direction of the major axis is specified.}
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{Show that the set of all points such that the difference of their}
\displaystyle \text{distances from }(4,0)\text{ and }(-4,0)\text{ is always equal to }2\text{ represents a hyperbola.}
\displaystyle \text{Answer:}
\displaystyle \text{The fixed points }(4,0)\text{ and }(-4,0)\text{ are the foci of the hyperbola.}
\displaystyle \therefore c=4.
\displaystyle \text{For a hyperbola, the difference of the distances from the foci is }2a.
\displaystyle \therefore 2a=2\Rightarrow a=1.
\displaystyle \therefore a^2=1.
\displaystyle \text{Also, }c^2=a^2+b^2.
\displaystyle 16=1+b^2.
\displaystyle \therefore b^2=15.
\displaystyle \text{Hence, the equation of the locus is}
\displaystyle \frac{x^2}{1}-\frac{y^2}{15}=1.
\displaystyle \therefore x^2-\frac{y^2}{15}=1.
\displaystyle \therefore \text{The given set of points represents a hyperbola.}
\displaystyle \\

\displaystyle \textbf{Question 58: }\text{The equations of two diameters of a circle are }x+y=6\text{ and }
\displaystyle x+2y=4, \ \text{and its radius is }10\text{ units. Find the equation of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{The centre of the circle is the point of intersection of the two diameters.}
\displaystyle x+y=6,\qquad x+2y=4.
\displaystyle \text{Subtracting the first equation from the second,}
\displaystyle y=-2.
\displaystyle \therefore x-2=6\Rightarrow x=8.
\displaystyle \therefore \text{Centre}=(8,-2).
\displaystyle \text{The equation of a circle with centre }(h,k)\text{ and radius }r\text{ is}
\displaystyle (x-h)^2+(y-k)^2=r^2.
\displaystyle \therefore (x-8)^2+(y+2)^2=100.
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{Find the equation of the circle which passes through the origin and}
\displaystyle \text{cuts off intercepts }4\text{ and }6\text{ on the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The general equation of a circle through the origin is}
\displaystyle x^2+y^2+Dx+Ey=0.
\displaystyle \text{Since the circle cuts an }x\text{-intercept }4,\text{ it passes through }(4,0).
\displaystyle 16+4D=0\Rightarrow D=-4.
\displaystyle \text{Since it cuts a }y\text{-intercept }6,\text{ it passes through }(0,6).
\displaystyle 36+6E=0\Rightarrow E=-6.
\displaystyle \therefore x^2+y^2-4x-6y=0.
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{If the length of the latus rectum of an ellipse with major axis along the}
\displaystyle x\text{-axis and centre at the origin is }20\text{ units, and the distance between the foci is equal}
\displaystyle \text{to the length of the minor axis, find the equation of the ellipse.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of the ellipse be }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a>b.
\displaystyle \text{Length of the latus rectum}=\frac{2b^2}{a}=20.
\displaystyle \therefore b^2=10a.\qquad\ldots(1)
\displaystyle \text{Distance between the foci}=2c\text{ and length of the minor axis}=2b.
\displaystyle \therefore 2c=2b\Rightarrow c=b.
\displaystyle \text{For an ellipse, }c^2=a^2-b^2.
\displaystyle \text{Since }c=b,
\displaystyle b^2=a^2-b^2.
\displaystyle \therefore a^2=2b^2.
\displaystyle \text{Using (1), }a^2=2(10a)=20a.
\displaystyle \therefore a=20.
\displaystyle \therefore a^2=400,\qquad b^2=200.
\displaystyle \therefore \frac{x^2}{400}+\frac{y^2}{200}=1.
\displaystyle \\

\displaystyle \textbf{Question 61: }\text{Find the equation of the ellipse whose foci are }(2,3)\text{ and }(-2,3)
\displaystyle \text{and whose semi-minor axis is }\sqrt5.
\displaystyle \text{Answer:}
\displaystyle \text{The centre is the midpoint of the foci.}
\displaystyle \therefore \text{Centre}=(0,3).
\displaystyle \text{Since the foci are }(0\pm2,3),\text{ we have }c=2.
\displaystyle \therefore c^2=4.
\displaystyle \text{Semi-minor axis }b=\sqrt5.
\displaystyle \therefore b^2=5.
\displaystyle \text{For an ellipse, }c^2=a^2-b^2.
\displaystyle 4=a^2-5.
\displaystyle \therefore a^2=9.
\displaystyle \text{Since the major axis is horizontal, the equation is}
\displaystyle \frac{x^2}{9}+\frac{(y-3)^2}{5}=1.
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{If the eccentricity of a hyperbola is }\sqrt3,\text{ find the eccentricity of}
\displaystyle \text{its conjugate hyperbola.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the hyperbola be }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.
\displaystyle \text{Its eccentricity is }e=\sqrt{1+\frac{b^2}{a^2}}.
\displaystyle \sqrt3=\sqrt{1+\frac{b^2}{a^2}}.
\displaystyle 3=1+\frac{b^2}{a^2}.
\displaystyle \therefore \frac{b^2}{a^2}=2.
\displaystyle \text{For the conjugate hyperbola, eccentricity }e'=\sqrt{1+\frac{a^2}{b^2}}.
\displaystyle \therefore e'=\sqrt{1+\frac{1}{2}}.
\displaystyle =\sqrt{\frac32}.
\displaystyle \therefore \text{The eccentricity of the conjugate hyperbola is }\sqrt{\frac32}.
\displaystyle \\

\displaystyle \textbf{Question 63: }\text{Find the vertex and focus of the parabola } \\ 2y^2+3y-4x-3=0.
\displaystyle \text{Answer:}
\displaystyle 2y^2+3y-4x-3=0.
\displaystyle 2y^2+3y=4x+3.
\displaystyle y^2+\frac{3}{2}y=2x+\frac{3}{2}.
\displaystyle y^2+\frac{3}{2}y+\frac{9}{16}=2x+\frac{3}{2}+\frac{9}{16}.
\displaystyle \left(y+\frac{3}{4}\right)^2=2x+\frac{33}{16}.
\displaystyle \left(y+\frac{3}{4}\right)^2=2\left(x+\frac{33}{32}\right).
\displaystyle \text{Comparing with }(y-k)^2=4a(x-h),
\displaystyle h=-\frac{33}{32},\qquad k=-\frac{3}{4},\qquad4a=2.
\displaystyle \therefore a=\frac{1}{2}.
\displaystyle \therefore \text{Vertex}=\left(-\frac{33}{32},-\frac{3}{4}\right).
\displaystyle \text{Focus}=(h+a,k).
\displaystyle =\left(-\frac{33}{32}+\frac{1}{2},-\frac{3}{4}\right).
\displaystyle \therefore \text{Focus}=\left(-\frac{17}{32},-\frac{3}{4}\right).
\displaystyle \\

\displaystyle \textbf{Question 64: }\text{Find the radius of a circle passing through the foci of the ellipse}
\displaystyle 9x^2+16y^2=144\text{ and having centre }(0,3).
\displaystyle \text{Answer:}
\displaystyle 9x^2+16y^2=144.
\displaystyle \therefore \frac{x^2}{16}+\frac{y^2}{9}=1.
\displaystyle \therefore a^2=16,\qquad b^2=9.
\displaystyle \text{For an ellipse, }c^2=a^2-b^2.
\displaystyle c^2=16-9=7.
\displaystyle \therefore c=\sqrt7.
\displaystyle \therefore \text{The foci are }(\pm\sqrt7,0).
\displaystyle \text{The centre of the required circle is }(0,3).
\displaystyle r=\sqrt{(\sqrt7-0)^2+(0-3)^2}.
\displaystyle =\sqrt{7+9}=\sqrt{16}=4.
\displaystyle \therefore \text{The radius of the circle is }4\text{ units.}
\displaystyle \\

\displaystyle \text{LONG ANSWER QUESTIONS}

\displaystyle \textbf{Question 65: }\text{Find the equation of the circle passing through the points }(-1,2)\text{ and}
\displaystyle (3,-2)\text{ and having its centre on the line }x-2y=0.
\displaystyle \text{Answer:}
\displaystyle \text{Let the centre of the circle be }C(h,k).
\displaystyle \text{Since }C\text{ lies on }x-2y=0,
\displaystyle h-2k=0.
\displaystyle \therefore h=2k.\qquad\ldots(1)
\displaystyle \text{Since }(-1,2)\text{ and }(3,-2)\text{ lie on the circle, their distances from }C
\displaystyle \text{are equal.}
\displaystyle (h+1)^2+(k-2)^2=(h-3)^2+(k+2)^2.
\displaystyle h^2+2h+1+k^2-4k+4=h^2-6h+9+k^2+4k+4.
\displaystyle 8h-8k-8=0.
\displaystyle \therefore h-k=1.\qquad\ldots(2)
\displaystyle \text{Using }h=2k\text{ in (2),}
\displaystyle 2k-k=1.
\displaystyle \therefore k=1\text{ and }h=2.
\displaystyle \therefore \text{The centre of the circle is }(2,1).
\displaystyle r^2=(-1-2)^2+(2-1)^2.
\displaystyle =9+1=10.
\displaystyle \therefore (x-2)^2+(y-1)^2=10.
\displaystyle x^2-4x+4+y^2-2y+1=10.
\displaystyle \therefore x^2+y^2-4x-2y-5=0.
\displaystyle \\

\displaystyle \textbf{Question 66: }\text{Prove that the points }(1,0),(2,-7),(8,1)\text{ and }(9,-6)\text{ are concyclic.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of the circle through }(1,0),(2,-7)\text{ and }(8,1)\text{ be}
\displaystyle x^2+y^2+Dx+Ey+F=0.
\displaystyle \text{Since }(1,0)\text{ lies on the circle,}
\displaystyle 1+D+F=0.\qquad\ldots(1)
\displaystyle \text{Since }(2,-7)\text{ lies on the circle,}
\displaystyle 4+49+2D-7E+F=0.
\displaystyle 53+2D-7E+F=0.\qquad\ldots(2)
\displaystyle \text{Since }(8,1)\text{ lies on the circle,}
\displaystyle 64+1+8D+E+F=0.
\displaystyle 65+8D+E+F=0.\qquad\ldots(3)
\displaystyle \text{Subtracting (1) from (2),}
\displaystyle D-7E+52=0.\qquad\ldots(4)
\displaystyle \text{Subtracting (1) from (3),}
\displaystyle 7D+E+64=0.\qquad\ldots(5)
\displaystyle \text{From (4), }D=7E-52.
\displaystyle \text{Substituting in (5),}
\displaystyle 7(7E-52)+E+64=0.
\displaystyle 50E-300=0.
\displaystyle \therefore E=6.
\displaystyle \therefore D=7(6)-52=-10.
\displaystyle \text{Using (1), }1-10+F=0.
\displaystyle \therefore F=9.
\displaystyle \therefore \text{The circle through the first three points is}
\displaystyle x^2+y^2-10x+6y+9=0.
\displaystyle \text{For the point }(9,-6),
\displaystyle 9^2+(-6)^2-10(9)+6(-6)+9
\displaystyle =81+36-90-36+9=0.
\displaystyle \therefore (9,-6)\text{ also lies on the same circle.}
\displaystyle \therefore \text{The four given points are concyclic.}
\displaystyle \\

\displaystyle \textbf{Question 67: }\text{Find the equation of the circle drawn on a diagonal of a rectangle as}
\displaystyle \text{diameter, whose sides are }x=4,\ x=-2,\ y=-3\text{ and }y=2.
\displaystyle \text{Answer:}
\displaystyle \text{The vertices of the rectangle are }(4,2),(4,-3),(-2,-3)\text{ and }(-2,2).
\displaystyle \text{Take the diagonal joining }A(4,2)\text{ and }C(-2,-3).
\displaystyle \text{The centre of the circle is the midpoint of }AC.
\displaystyle \therefore \text{Centre}=\left(\frac{4+(-2)}{2},\frac{2+(-3)}{2}\right).
\displaystyle =\left(1,-\frac{1}{2}\right).
\displaystyle AC^2=(4+2)^2+(2+3)^2.
\displaystyle =6^2+5^2=61.
\displaystyle \therefore r^2=\frac{AC^2}{4}=\frac{61}{4}.
\displaystyle \text{Hence, the equation of the circle is}
\displaystyle (x-1)^2+\left(y+\frac{1}{2}\right)^2=\frac{61}{4}.
\displaystyle x^2-2x+1+y^2+y+\frac{1}{4}=\frac{61}{4}.
\displaystyle \therefore x^2+y^2-2x+y-14=0.
\displaystyle \\

\displaystyle \textbf{Question 68: }\text{Find the equation of a circle which touches the }y\text{-axis at the point }(0,4)
\displaystyle \text{and cuts off an intercept of length }6\text{ on the }x\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the centre of the circle be }C(h,4).
\displaystyle \text{Since the circle touches the }y\text{-axis at }(0,4),
\displaystyle r=|h|.
\displaystyle \therefore \text{The equation of the circle is}
\displaystyle (x-h)^2+(y-4)^2=h^2.
\displaystyle \text{Putting }y=0\text{ to find the points where it meets the }x\text{-axis,}
\displaystyle (x-h)^2+16=h^2.
\displaystyle \therefore (x-h)^2=h^2-16.
\displaystyle \therefore x=h\pm\sqrt{h^2-16}.
\displaystyle \text{The length of the intercept on the }x\text{-axis is}
\displaystyle 2\sqrt{h^2-16}=6.
\displaystyle \sqrt{h^2-16}=3.
\displaystyle h^2-16=9.
\displaystyle \therefore h^2=25\Rightarrow h=\pm5.
\displaystyle \text{For }h=5,\text{ the circle is}
\displaystyle (x-5)^2+(y-4)^2=25.
\displaystyle \therefore x^2+y^2-10x-8y+16=0.
\displaystyle \text{For }h=-5,\text{ the circle is}
\displaystyle (x+5)^2+(y-4)^2=25.
\displaystyle \therefore x^2+y^2+10x-8y+16=0.
\displaystyle \therefore \text{These are the two possible equations of the circle.}
\displaystyle \\

\displaystyle \textbf{Question 69: }\text{Find the equation of the circle which passes through the points }
\displaystyle (0,2),(3,0) \ \text{and }(3,2).
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of the circle be}
\displaystyle x^2+y^2+Dx+Ey+F=0.
\displaystyle \text{Since }(0,2)\text{ lies on the circle,}
\displaystyle 4+2E+F=0.\qquad\ldots(1)
\displaystyle \text{Since }(3,0)\text{ lies on the circle,}
\displaystyle 9+3D+F=0.\qquad\ldots(2)
\displaystyle \text{Since }(3,2)\text{ lies on the circle,}
\displaystyle 13+3D+2E+F=0.\qquad\ldots(3)
\displaystyle \text{Subtracting (2) from (3),}
\displaystyle 4+2E=0.
\displaystyle \therefore E=-2.
\displaystyle \text{Substituting }E=-2\text{ in (1),}
\displaystyle 4-4+F=0.
\displaystyle \therefore F=0.
\displaystyle \text{Substituting }F=0\text{ in (2),}
\displaystyle 9+3D=0.
\displaystyle \therefore D=-3.
\displaystyle \therefore x^2+y^2-3x-2y=0.
\displaystyle \\

\displaystyle \textbf{Question 70: }\text{Find the equation of the circle which passes through the centre of the circle}
\displaystyle x^2+y^2+8x+10y-7=0\text{ and is concentric with the circle}
\displaystyle 2x^2+2y^2-8x-12y-9=0.
\displaystyle \text{Answer:}
\displaystyle x^2+y^2+8x+10y-7=0.
\displaystyle \text{Its centre is }\left(-\frac{8}{2},-\frac{10}{2}\right)=(-4,-5).
\displaystyle 2x^2+2y^2-8x-12y-9=0.
\displaystyle \text{Dividing by }2,
\displaystyle x^2+y^2-4x-6y-\frac{9}{2}=0.
\displaystyle \therefore \text{Its centre is }(2,3).
\displaystyle \text{The required circle is concentric with this circle, so its centre is }(2,3).
\displaystyle \text{It passes through }(-4,-5).
\displaystyle r^2=(-4-2)^2+(-5-3)^2.
\displaystyle =(-6)^2+(-8)^2=36+64=100.
\displaystyle \therefore (x-2)^2+(y-3)^2=100.
\displaystyle \therefore x^2+y^2-4x-6y-87=0.
\displaystyle \\

\displaystyle \textbf{Question 71: }\text{An equilateral triangle is inscribed in the parabola }y^2=4ax,\text{ with one}
\displaystyle \text{vertex at the vertex of the parabola. Find the length of the side of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }O(0,0)\text{ be one vertex of the equilateral triangle.}
\displaystyle \text{By symmetry, let the other two vertices be }P(x,y)\text{ and }Q(x,-y).
\displaystyle \text{Since the triangle is equilateral, }\angle POQ=60^\circ.
\displaystyle \therefore \angle POX=30^\circ.
\displaystyle \therefore \tan30^\circ=\frac{y}{x}.
\displaystyle \frac{1}{\sqrt3}=\frac{y}{x}.
\displaystyle \therefore y=\frac{x}{\sqrt3}.
\displaystyle \text{Since }P(x,y)\text{ lies on }y^2=4ax,
\displaystyle \frac{x^2}{3}=4ax.
\displaystyle \therefore x=12a.
\displaystyle \therefore y=\frac{12a}{\sqrt3}=4\sqrt3a.
\displaystyle \text{The side }PQ=2y.
\displaystyle \therefore PQ=8\sqrt3a.
\displaystyle \therefore \text{The length of the side of the triangle is }8\sqrt3a.
\displaystyle \\

\displaystyle \textbf{Question 72: }\text{A rod of length }12\text{ cm moves with its ends always touching the coordinate axes.}
\displaystyle \text{Determine the locus of a point }P\text{ on the rod which is }3\text{ cm from the end in contact}
\displaystyle \text{with the }x\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the ends of the rod be }A(a,0)\text{ and }B(0,b).
\displaystyle \text{Since the length of the rod is }12\text{ cm,}
\displaystyle a^2+b^2=144.\qquad\ldots(1)
\displaystyle \text{The point }P\text{ is }3\text{ cm from }A.
\displaystyle \therefore AP=3,\qquad PB=9.
\displaystyle \therefore AP:PB=1:3.
\displaystyle \text{Let }P(x,y)\text{ divide }AB\text{ internally in the ratio }1:3.
\displaystyle \therefore x=\frac{3a}{4},\qquad y=\frac{b}{4}.
\displaystyle \therefore a=\frac{4x}{3},\qquad b=4y.
\displaystyle \text{Substituting in (1),}
\displaystyle \left(\frac{4x}{3}\right)^2+(4y)^2=144.
\displaystyle \frac{16x^2}{9}+16y^2=144.
\displaystyle \therefore \frac{x^2}{9}+y^2=9.
\displaystyle \therefore \frac{x^2}{81}+\frac{y^2}{9}=1.
\displaystyle \therefore \text{The locus of }P\text{ is the ellipse }\frac{x^2}{81}+\frac{y^2}{9}=1.
\displaystyle \\

\displaystyle \textbf{Question 73: }\text{If the lines }2x-3y=5\text{ and }3x-4y=7\text{ are the diameters of a circle}
\displaystyle \text{of area }154\text{ square units, obtain the equation of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{The centre of the circle is the point of intersection of the two diameters.}
\displaystyle 2x-3y=5.\qquad\ldots(1)
\displaystyle 3x-4y=7.\qquad\ldots(2)
\displaystyle \text{Multiplying (1) by }4\text{ and (2) by }3,
\displaystyle 8x-12y=20,
\displaystyle 9x-12y=21.
\displaystyle \text{Subtracting, }x=1.
\displaystyle \text{Substituting }x=1\text{ in (1),}
\displaystyle 2-3y=5.
\displaystyle \therefore y=-1.
\displaystyle \therefore \text{Centre}=(1,-1).
\displaystyle \text{Area of the circle}=\pi r^2=154.
\displaystyle \frac{22}{7}r^2=154.
\displaystyle \therefore r^2=49\Rightarrow r=7.
\displaystyle \text{Hence, the equation of the circle is}
\displaystyle (x-1)^2+(y+1)^2=49.
\displaystyle \\

\displaystyle \textbf{Question 74: }\text{Find the equation of the set of all points the sum of whose distances}
\displaystyle \text{from the points }(3,0)\text{ and }(9,0)\text{ is }12.
\displaystyle \text{Answer:}
\displaystyle \text{The locus is an ellipse with foci }(3,0)\text{ and }(9,0).
\displaystyle \text{The centre is the midpoint of the foci.}
\displaystyle \therefore \text{Centre}=\left(\frac{3+9}{2},0\right)=(6,0).
\displaystyle \text{Distance between the foci}=2c=9-3=6.
\displaystyle \therefore c=3.
\displaystyle \text{The sum of the distances from the foci is }2a=12.
\displaystyle \therefore a=6.
\displaystyle \text{For an ellipse, }c^2=a^2-b^2.
\displaystyle 3^2=6^2-b^2.
\displaystyle 9=36-b^2.
\displaystyle \therefore b^2=27.
\displaystyle \text{Since the major axis is parallel to the }x\text{-axis, the equation is}
\displaystyle \frac{(x-6)^2}{36}+\frac{y^2}{27}=1.
\displaystyle \\

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