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

\displaystyle \textbf{Question 1: }\text{The coordinates of the focus of the parabola}
\displaystyle y^2-x-2y+2=0\text{ are}
\displaystyle \text{(a) }\left(\frac54,1\right)\qquad\text{(b) }\left(\frac14,0\right)\qquad\text{(c) }(1,1)\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle y^2-2y=x-2\Rightarrow(y-1)^2=x-1.
\displaystyle \text{Comparing with }(y-k)^2=4a(x-h),\quad4a=1\Rightarrow a=\frac14.
\displaystyle \text{Focus}=(h+a,k)=\left(1+\frac14,1\right)=\left(\frac54,1\right).
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\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) none of these}
\displaystyle \text{Answer:}
\displaystyle (y+a)^2=8a(x-a).
\displaystyle \text{Comparing with }(y-k)^2=4A(x-h),\quad h=a,\ k=-a.
\displaystyle \therefore\text{Vertex}=(a,-a).
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the focus of a parabola is }(-2,1)\text{ and the}
\displaystyle \text{directrix has the equation }x+y=3,\text{ then its vertex is}
\displaystyle \text{(a) }(0,3)\qquad\text{(b) }\left(-1,\frac12\right)\qquad\text{(c) }(-1,2)\qquad\text{(d) }(2,-1)
\displaystyle \text{Answer:}
\displaystyle \text{The foot of the perpendicular from }(-2,1)\text{ to }x+y=3\text{ is }(0,3).
\displaystyle \text{The vertex is the midpoint of the focus and this foot.}
\displaystyle V=\left(\frac{-2+0}{2},\frac{1+3}{2}\right)=(-1,2).
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The equation of the parabola whose vertex is }(a,0)
\displaystyle \text{and the directrix has the equation }x+y=3a,\text{ is}
\displaystyle \text{(a) }x^2+y^2+2xy+6ax+10ay+7a^2=0
\displaystyle \text{(b) }x^2-2xy+y^2+6ax+10ay-7a^2=0
\displaystyle \text{(c) }x^2-2xy+y^2-6ax+10ay-7a^2=0
\displaystyle \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{The foot from }(a,0)\text{ to }x+y=3a\text{ is }(2a,a).
\displaystyle \therefore\text{Focus}=2(a,0)-(2a,a)=(0,-a).
\displaystyle \sqrt{x^2+(y+a)^2}=\frac{|x+y-3a|}{\sqrt2}.
\displaystyle 2\{x^2+(y+a)^2\}=(x+y-3a)^2.
\displaystyle \therefore x^2-2xy+y^2+6ax+10ay-7a^2=0.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The parametric equations of a parabola are }x=t^2+1,
\displaystyle y=2t+1.\text{ The cartesian equation of its directrix is}
\displaystyle \text{(a) }x=0\qquad\text{(b) }x+1=0\qquad\text{(c) }y=0\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle y-1=2t\Rightarrow t=\frac{y-1}{2}.
\displaystyle x-1=t^2=\frac{(y-1)^2}{4}.
\displaystyle \therefore (y-1)^2=4(x-1).
\displaystyle \text{Here }a=1,\ h=1.\quad\therefore\text{Directrix: }x=h-a=0.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If the coordinates of the vertex and the focus of a}
\displaystyle \text{parabola are }(-1,1)\text{ and }(2,3)\text{ respectively, then the equation}
\displaystyle \text{of its directrix is}
\displaystyle \text{(a) }3x+2y+14=0\qquad\text{(b) }3x+2y-25=0
\displaystyle \text{(c) }2x-3y+10=0\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle V=(-1,1),\quad S=(2,3)\Rightarrow\overrightarrow{VS}=(3,2).
\displaystyle \text{The point on the directrix along the axis is }D=2V-S=(-4,-1).
\displaystyle \text{The directrix has normal }(3,2)\text{ and passes through }(-4,-1).
\displaystyle 3(x+4)+2(y+1)=0\Rightarrow3x+2y+14=0.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The locus of the points of trisection of the double}
\displaystyle \text{ordinates of a parabola is a}
\displaystyle \text{(a) pair of lines}\qquad\text{(b) circle}\qquad\text{(c) parabola}\qquad\text{(d) straight line}
\displaystyle \text{Answer:}
\displaystyle \text{Let the parabola be }y^2=4ax.
\displaystyle \text{A double ordinate has endpoints }(x,y)\text{ and }(x,-y).
\displaystyle \text{A point of trisection is }\left(x,\frac{y}{3}\right).
\displaystyle \text{Let }Y=\frac{y}{3},\ X=x.\text{ Then }y=3Y.
\displaystyle 9Y^2=4aX,\text{ which represents a parabola.}
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The equation of the directrix of the parabola whose}
\displaystyle \text{vertex and focus are }(1,4)\text{ and }(2,6)\text{ respectively is}
\displaystyle \text{(a) }x+2y=4\qquad\text{(b) }x-y=3\qquad\text{(c) }2x+y=5\qquad\text{(d) }x+3y=8
\displaystyle \text{Answer:}
\displaystyle V=(1,4),\quad S=(2,6)\Rightarrow\overrightarrow{VS}=(1,2).
\displaystyle \text{The corresponding point on the directrix is }D=2V-S=(0,2).
\displaystyle \text{The directrix has normal }(1,2)\text{ and passes through }(0,2).
\displaystyle x+2(y-2)=0\Rightarrow x+2y=4.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }V\text{ and }S\text{ are respectively the vertex and focus}
\displaystyle \text{of the parabola }y^2+6y+2x+5=0,\text{ then }SV=
\displaystyle \text{(a) }2\qquad\text{(b) }\frac12\qquad\text{(c) }1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle y^2+6y+2x+5=0\Rightarrow(y+3)^2=-2(x-2).
\displaystyle \text{Comparing with }(y-k)^2=-4a(x-h),\quad4a=2.
\displaystyle \therefore a=\frac12.
\displaystyle \text{Since }SV=a,\quad SV=\frac12.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The directrix of the parabola}
\displaystyle x^2-4x-8y+12=0\text{ is}
\displaystyle \text{(a) }y=0\qquad\text{(b) }x=1\qquad\text{(c) }y=-1\qquad\text{(d) }x=-1
\displaystyle \text{Answer:}
\displaystyle x^2-4x-8y+12=0\Rightarrow(x-2)^2=8(y-1).
\displaystyle \text{Comparing with }(x-h)^2=4a(y-k),\quad a=2,\ k=1.
\displaystyle \therefore\text{Directrix: }y=k-a=1-2=-1.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The equation of the parabola with focus }(0,0)
\displaystyle \text{and directrix }x+y=4\text{ is}
\displaystyle \text{(a) }x^2+y^2-2xy+8x+8y-16=0
\displaystyle \text{(b) }x^2+y^2-2xy+8x+8y=0
\displaystyle \text{(c) }x^2+y^2+8x+8y-16=0
\displaystyle \text{(d) }x^2-y^2+8x+8y-16=0
\displaystyle \text{Answer:}
\displaystyle \sqrt{x^2+y^2}=\frac{|x+y-4|}{\sqrt2}.
\displaystyle 2(x^2+y^2)=(x+y-4)^2.
\displaystyle \therefore x^2+y^2-2xy+8x+8y-16=0.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The line }2x-y+4=0\text{ cuts the parabola }y^2=8x
\displaystyle \text{in }P\text{ and }Q.\text{ The mid-point of }PQ\text{ is}
\displaystyle \text{(a) }(1,2)\qquad\text{(b) }(1,-2)\qquad\text{(c) }(-1,2)\qquad\text{(d) }(-1,-2)
\displaystyle \text{Answer:}
\displaystyle 2x-y+4=0\Rightarrow y=2x+4\Rightarrow x=\frac{y-4}{2}.
\displaystyle y^2=8\left(\frac{y-4}{2}\right)\Rightarrow y^2-4y+16=0.
\displaystyle y_1+y_2=4\Rightarrow y_M=\frac{y_1+y_2}{2}=2.
\displaystyle 2x_M-y_M+4=0\Rightarrow2x_M-2+4=0\Rightarrow x_M=-1.
\displaystyle \therefore M=(-1,2).
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In the parabola }y^2=4ax,\text{ the length of the chord}
\displaystyle \text{passing through the vertex and inclined to the axis at }\pi/4\text{ is}
\displaystyle \text{(a) }4\sqrt2\,a\qquad\text{(b) }2\sqrt2\,a\qquad\text{(c) }\sqrt2\,a\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{The chord has slope }\tan\frac{\pi}{4}=1,\text{ so its equation is }y=x.
\displaystyle x^2=4ax\Rightarrow x=0\text{ or }x=4a.
\displaystyle \text{The other end of the chord is }(4a,4a).
\displaystyle \therefore\text{Chord length}=\sqrt{(4a)^2+(4a)^2}=4\sqrt2\,a.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The equation }16x^2+y^2+8xy-74x-78y+212=0
\displaystyle \text{represents}
\displaystyle \text{(a) a circle}\qquad\text{(b) a parabola}\qquad\text{(c) an ellipse}\qquad\text{(d) a hyperbola}
\displaystyle \text{Answer:}
\displaystyle \text{For }Ax^2+Bxy+Cy^2+\cdots=0,\quad A=16,\ B=8,\ C=1.
\displaystyle B^2-4AC=8^2-4(16)(1)=64-64=0.
\displaystyle \therefore\text{The equation represents a parabola.}
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The length of the latus-rectum of the parabola}
\displaystyle y^2+8x-2y+17=0\text{ is}
\displaystyle \text{(a) }2\qquad\text{(b) }4\qquad\text{(c) }8\qquad\text{(d) }16
\displaystyle \text{Answer:}
\displaystyle y^2-2y+8x+17=0\Rightarrow(y-1)^2=-8(x+2).
\displaystyle \text{Comparing with }(y-k)^2=-4a(x-h),\quad4a=8.
\displaystyle \therefore\text{Length of latus-rectum}=4a=8.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The vertex of the parabola }x^2+8x+12y+4=0\text{ is}
\displaystyle \text{(a) }(-4,1)\qquad\text{(b) }(4,-1)\qquad\text{(c) }(-4,-1)\qquad\text{(d) }(4,1)
\displaystyle \text{Answer:}
\displaystyle x^2+8x+12y+4=0.
\displaystyle (x+4)^2=-12(y-1).
\displaystyle \therefore\text{Vertex}=(-4,1).
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The vertex of the parabola }(y-2)^2=16(x-1)\text{ is}
\displaystyle \text{(a) }(1,2)\qquad\text{(b) }(-1,2)\qquad\text{(c) }(1,-2)\qquad\text{(d) }(2,1)
\displaystyle \text{Answer:}
\displaystyle \text{Comparing with }(y-k)^2=4a(x-h),\quad h=1,\ k=2.
\displaystyle \therefore\text{Vertex}=(1,2).
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The length of the latus-rectum of the parabola}
\displaystyle 4y^2+2x-20y+17=0\text{ is}
\displaystyle \text{(a) }3\qquad\text{(b) }6\qquad\text{(c) }\frac12\qquad\text{(d) }9
\displaystyle \text{Answer:}
\displaystyle 4(y^2-5y)+2x+17=0.
\displaystyle 4\left(y-\frac52\right)^2=-2(x-4).
\displaystyle \left(y-\frac52\right)^2=-\frac12(x-4).
\displaystyle \text{Comparing with }(y-k)^2=-4a(x-h),\quad4a=\frac12.
\displaystyle \therefore\text{Length of latus-rectum}=4a=\frac12.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The length of the latus-rectum of the parabola}
\displaystyle x^2-4x-8y+12=0\text{ is}
\displaystyle \text{(a) }4\qquad\text{(b) }6\qquad\text{(c) }8\qquad\text{(d) }10
\displaystyle \text{Answer:}
\displaystyle x^2-4x-8y+12=0\Rightarrow(x-2)^2=8(y-1).
\displaystyle \text{Comparing with }(x-h)^2=4a(y-k),\quad4a=8.
\displaystyle \therefore\text{Length of latus-rectum}=8.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The focus of the parabola }y=2x^2+x\text{ is}
\displaystyle \text{(a) }(0,0)\qquad\text{(b) }\left(\frac12,\frac14\right)\qquad\text{(c) }\left(-\frac14,0\right)\qquad\text{(d) }\left(-\frac14,\frac18\right)
\displaystyle \text{Answer:}
\displaystyle y=2x^2+x=2\left(x+\frac14\right)^2-\frac18.
\displaystyle \left(x+\frac14\right)^2=\frac12\left(y+\frac18\right).
\displaystyle \text{Thus, }4a=\frac12\Rightarrow a=\frac18,\quad\text{vertex}=\left(-\frac14,-\frac18\right).
\displaystyle \therefore\text{Focus}=\left(-\frac14,-\frac18+\frac18\right)=\left(-\frac14,0\right).
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Which of the following points lie on the parabola }x^2=4ay?
\displaystyle \text{(a) }x=at^2,\ y=2at\qquad\text{(b) }x=2at,\ y=at^2
\displaystyle \text{(c) }x=2at^2,\ y=at\qquad\text{(d) }x=2at,\ y=at^2
\displaystyle \text{Answer:}
\displaystyle \text{For }x^2=4ay,\text{ put }y=at^2.
\displaystyle x^2=4a^2t^2\Rightarrow x=2at.
\displaystyle \therefore (x,y)=(2at,at^2).
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The equation of the parabola whose focus is }(1,-1)
\displaystyle \text{and the directrix is }x+y+7=0\text{ is}
\displaystyle \text{(a) }x^2+y^2-2xy-18x-10y=0
\displaystyle \text{(b) }x^2-18x-10y-45=0
\displaystyle \text{(c) }x^2+y^2-18x-10y-45=0
\displaystyle \text{(d) }x^2+y^2-2xy-18x-10y-45=0
\displaystyle \text{Answer:}
\displaystyle \text{For }P(x,y),\quad SP=\frac{|x+y+7|}{\sqrt2}.
\displaystyle (x-1)^2+(y+1)^2=\frac{(x+y+7)^2}{2}.
\displaystyle 2\{(x-1)^2+(y+1)^2\}=(x+y+7)^2.
\displaystyle \therefore x^2+y^2-2xy-18x-10y-45=0.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{For the ellipse }12x^2+4y^2+24x-16y+25=0
\displaystyle \text{(a) centre is }(-1,2)\qquad\text{(b) lengths of the axes are }\sqrt3\text{ and }1
\displaystyle \text{(c) eccentricity}=\sqrt{\frac23}\qquad\text{(d) all of these}
\displaystyle \text{Answer:}
\displaystyle 12(x+1)^2+4(y-2)^2=3.
\displaystyle \therefore\frac{(x+1)^2}{1/4}+\frac{(y-2)^2}{3/4}=1.
\displaystyle \text{Centre}=(-1,2),\quad a^2=\frac34,\quad b^2=\frac14.
\displaystyle 2a=\sqrt3,\qquad2b=1.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac13}=\sqrt{\frac23}.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The equation of the ellipse with focus }(-1,1),
\displaystyle \text{directrix }x-y+3=0\text{ and eccentricity }\frac12\text{ is}
\displaystyle \text{(a) }7x^2+2xy+7y^2+10x+10y+7=0
\displaystyle \text{(b) }7x^2+2xy+7y^2+10x-10y+7=0
\displaystyle \text{(c) }7x^2+2xy+7y^2+10x-10y-7=0
\displaystyle \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle (x+1)^2+(y-1)^2 =\frac14\left(\frac{x-y+3}{\sqrt2}\right)^2.
\displaystyle 8\{(x+1)^2+(y-1)^2\}=(x-y+3)^2.
\displaystyle \therefore7x^2+2xy+7y^2+10x-10y+7=0.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{The equation of the circle drawn with the two foci of}
\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ as the end-points of a diameter is}
\displaystyle \text{(a) }x^2+y^2=a^2+b^2\qquad\text{(b) }x^2+y^2=a^2
\displaystyle \text{(c) }x^2+y^2=2a^2\qquad\text{(d) }x^2+y^2=a^2-b^2
\displaystyle \text{Answer:}
\displaystyle \text{The foci are }(ae,0)\text{ and }(-ae,0).
\displaystyle \text{Hence, the circle has centre }(0,0)\text{ and radius }ae.
\displaystyle r^2=a^2e^2=a^2-b^2.
\displaystyle \therefore x^2+y^2=a^2-b^2.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{The eccentricity of the ellipse } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1
\displaystyle \text{if its latus-rectum is equal to one half of its minor axis, is}
\displaystyle \text{(a) }\frac1{\sqrt2}\qquad\text{(b) }\frac{\sqrt3}{2} \qquad\text{(c) }\frac12\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{2b^2}{a}=\frac12(2b)\Rightarrow\frac{2b^2}{a}=b.
\displaystyle \therefore\frac ba=\frac12.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac14}=\frac{\sqrt3}{2}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{The eccentricity of the ellipse, if the distance}
\displaystyle \text{between the foci is equal to the length of the latus-rectum, is}
\displaystyle \text{(a) }\frac{\sqrt5-1}{2}\qquad\text{(b) }\frac{\sqrt5+1}{2} \qquad\text{(c) }\frac{\sqrt5-1}{4}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 2ae=\frac{2b^2}{a}\Rightarrow e=\frac{b^2}{a^2}=1-e^2.
\displaystyle e^2+e-1=0.
\displaystyle \therefore e=\frac{-1+\sqrt5}{2}=\frac{\sqrt5-1}{2}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{The eccentricity of the ellipse, if the minor axis}
\displaystyle \text{is equal to the distance between the foci, is}
\displaystyle \text{(a) }\frac{\sqrt3}{2}\qquad\text{(b) }\frac2{\sqrt3} \qquad\text{(c) }\frac1{\sqrt2}\qquad\text{(d) }\frac{\sqrt2}{3}
\displaystyle \text{Answer:}
\displaystyle \text{Minor axis}=2b,\qquad\text{distance between foci}=2ae.
\displaystyle 2b=2ae\Rightarrow\frac ba=e.
\displaystyle e^2=1-\frac{b^2}{a^2}=1-e^2.
\displaystyle 2e^2=1\Rightarrow e=\frac1{\sqrt2}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{The difference between the lengths of the major axis}
\displaystyle \text{and the latus-rectum of an ellipse is}
\displaystyle \text{(a) }ae\qquad\text{(b) }2ae\qquad\text{(c) }ae^2 \qquad\text{(d) }2ae^2
\displaystyle \text{Answer:}
\displaystyle \text{Major axis}=2a,\qquad\text{latus-rectum}=\frac{2b^2}{a}.
\displaystyle \text{Difference}=2a-\frac{2b^2}{a} =2a\left(1-\frac{b^2}{a^2}\right).
\displaystyle \text{Since }e^2=1-\frac{b^2}{a^2},\quad \text{difference}=2ae^2.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{The eccentricity of the conic }9x^2+25y^2=225\text{ is}
\displaystyle \text{(a) }\frac25\qquad\text{(b) }\frac45\qquad \text{(c) }\frac13\qquad\text{(d) }\frac15
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{25}+\frac{y^2}{9}=1 \Rightarrow a^2=25,\quad b^2=9.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac9{25}}=\frac45.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{The latus-rectum of the conic}
\displaystyle 3x^2+4y^2-6x+8y-5=0\text{ is}
\displaystyle \text{(a) }3\qquad\text{(b) }\frac{\sqrt3}{2}\qquad \text{(c) }\frac{2}{\sqrt3}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 3(x-1)^2+4(y+1)^2=12.
\displaystyle \therefore\frac{(x-1)^2}{4}+\frac{(y+1)^2}{3}=1.
\displaystyle a=2,\quad b^2=3.
\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a} =\frac{2(3)}{2}=3.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{The equations of the tangents to the ellipse}
\displaystyle 9x^2+16y^2=144\text{ from the point }(2,3)\text{ are}
\displaystyle \text{(a) }y=3,\ x=5\qquad\text{(b) }x=2,\ y=3
\displaystyle \text{(c) }x=3,\ y=2\qquad\text{(d) }x+y=5,\ y=3
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{16}+\frac{y^2}{9}=1.
\displaystyle \text{Let a tangent through }(2,3)\text{ be }y=mx+c.
\displaystyle c=3-2m.
\displaystyle \text{For tangency, }c^2=16m^2+9.
\displaystyle (3-2m)^2=16m^2+9\Rightarrow m(m+1)=0.
\displaystyle m=0\Rightarrow y=3.
\displaystyle m=-1\Rightarrow y-3=-(x-2)\Rightarrow x+y=5.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{The eccentricity of the ellipse}
\displaystyle 4x^2+9y^2+8x+36y+4=0\text{ is}
\displaystyle \text{(a) }\frac56\qquad\text{(b) }\frac35\qquad \text{(c) }\frac{\sqrt2}{3}\qquad\text{(d) }\frac{\sqrt5}{3}
\displaystyle \text{Answer:}
\displaystyle 4(x^2+2x)+9(y^2+4y)+4=0.
\displaystyle 4(x+1)^2+9(y+2)^2=36.
\displaystyle \therefore\frac{(x+1)^2}{9}+\frac{(y+2)^2}{4}=1.
\displaystyle a^2=9,\quad b^2=4.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac49}=\frac{\sqrt5}{3}.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{The eccentricity of the ellipse } 4x^2+9y^2=36\text{ is}
\displaystyle \text{(a) }\frac1{2\sqrt3}\qquad\text{(b) }\frac1{\sqrt3} \qquad\text{(c) }\frac{\sqrt5}{3}\qquad\text{(d) }\frac{\sqrt5}{6}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1\Rightarrow a^2=9,\quad b^2=4.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac49}=\frac{\sqrt5}{3}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{The eccentricity of the ellipse } 5x^2+9y^2=1\text{ is}
\displaystyle \text{(a) }\frac23\qquad\text{(b) }\frac34\qquad \text{(c) }\frac45\qquad\text{(d) }\frac12
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{1/5}+\frac{y^2}{1/9}=1 \Rightarrow a^2=\frac15,\quad b^2=\frac19.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac59}=\frac23.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{For the ellipse }x^2+4y^2=9
\displaystyle \text{(a) the eccentricity is }\frac12 \qquad\text{(b) the latus-rectum is }\frac32
\displaystyle \text{(c) a focus is }(3\sqrt3,0) \qquad\text{(d) a directrix is }x=-2\sqrt3
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{9}+\frac{y^2}{9/4}=1 \Rightarrow a=3,\quad b=\frac32.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac14}=\frac{\sqrt3}{2}.
\displaystyle \text{Latus-rectum}=\frac{2b^2}{a} =\frac{2(9/4)}{3}=\frac32.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{If the latus-rectum of an ellipse is one half}
\displaystyle \text{of its minor axis, then its eccentricity is}
\displaystyle \text{(a) }\frac12\qquad\text{(b) }\frac1{\sqrt2}\qquad \text{(c) }\frac{\sqrt3}{2}\qquad\text{(d) }\frac{\sqrt3}{4}
\displaystyle \text{Answer:}
\displaystyle \frac{2b^2}{a}=\frac12(2b)\Rightarrow\frac ba=\frac12.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac14}=\frac{\sqrt3}{2}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{An ellipse has its centre at }(1,-1)\text{ and}
\displaystyle \text{semi-major axis }=8\text{ and it passes through }(1,3).\text{ The equation}
\displaystyle \text{of the ellipse is}
\displaystyle \text{(a) }\frac{(x+1)^2}{64}+\frac{(y+1)^2}{16}=1\qquad \text{(b) }\frac{(x-1)^2}{64}+\frac{(y+1)^2}{16}=1
\displaystyle \text{(c) }\frac{(x-1)^2}{16}+\frac{(y+1)^2}{64}=1\qquad \text{(d) }\frac{(x+1)^2}{64}+\frac{(y-1)^2}{16}=1
\displaystyle \text{Answer:}
\displaystyle \text{Centre}=(1,-1),\quad a=8\Rightarrow a^2=64.
\displaystyle \text{Let }\frac{(x-1)^2}{64}+\frac{(y+1)^2}{b^2}=1.
\displaystyle \text{Since }(1,3)\text{ lies on it, }\frac{(3+1)^2}{b^2}=1.
\displaystyle \therefore b^2=16.
\displaystyle \therefore\frac{(x-1)^2}{64}+\frac{(y+1)^2}{16}=1.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{The sum of the focal distances of any point on}
\displaystyle \text{the ellipse }9x^2+16y^2=144\text{ is}
\displaystyle \text{(a) }32\qquad\text{(b) }18\qquad \text{(c) }16\qquad\text{(d) }8
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{16}+\frac{y^2}{9}=1\Rightarrow a=4.
\displaystyle \text{Sum of focal distances}=2a=8.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{If }(2,4)\text{ and }(10,10)\text{ are the ends of a}
\displaystyle \text{latus-rectum of an ellipse with eccentricity }\frac12,\text{ then the length}
\displaystyle \text{of semi-major axis is}
\displaystyle \text{(a) }\frac{20}{3}\qquad\text{(b) }\frac{15}{3}\qquad \text{(c) }\frac{40}{3}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Latus-rectum}=\sqrt{(10-2)^2+(10-4)^2}=10.
\displaystyle e=\frac12\Rightarrow b^2=a^2(1-e^2)=\frac34a^2.
\displaystyle \frac{2b^2}{a}=10\Rightarrow \frac{2(3a^2/4)}{a}=10.
\displaystyle \frac{3a}{2}=10\Rightarrow a=\frac{20}{3}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{The equation } \frac{x^2}{2-\lambda}+\frac{y^2}{\lambda-5}+1=0\text{ represents an ellipse, if}
\displaystyle \text{(a) }\lambda<5\qquad\text{(b) }\lambda<2\qquad \text{(c) }2<\lambda<5\qquad\text{(d) }\lambda<2\text{ or }\lambda>5
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{\lambda-2}+\frac{y^2}{5-\lambda}=1.
\displaystyle \text{For an ellipse, }\lambda-2>0\text{ and }5-\lambda>0.
\displaystyle \therefore\lambda>2\text{ and }\lambda<5\Rightarrow2<\lambda<5.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{The eccentricity of the ellipse}
\displaystyle 9x^2+25y^2-18x-100y-116=0\text{ is}
\displaystyle \text{(a) }\frac{25}{16}\qquad\text{(b) }\frac45\qquad \text{(c) }\frac{16}{25}\qquad\text{(d) }\frac54
\displaystyle \text{Answer:}
\displaystyle 9(x-1)^2+25(y-2)^2=225.
\displaystyle \therefore\frac{(x-1)^2}{25}+\frac{(y-2)^2}{9}=1.
\displaystyle a^2=25,\quad b^2=9.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac9{25}}=\frac45.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{If the major axis of an ellipse is three times}
\displaystyle \text{the minor axis, then its eccentricity is equal to}
\displaystyle \text{(a) }\frac13\qquad\text{(b) }\frac1{\sqrt3}\qquad \text{(c) }\frac1{\sqrt2}\qquad\text{(d) }\frac{2\sqrt2}{3}
\displaystyle \text{Answer:}
\displaystyle 2a=3(2b)\Rightarrow a=3b\Rightarrow\frac{b^2}{a^2}=\frac19.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac19}=\frac{2\sqrt2}{3}.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{The eccentricity of the ellipse } 25x^2+16y^2=400\text{ is}
\displaystyle \text{(a) }\frac35\qquad\text{(b) }\frac13\qquad \text{(c) }\frac25\qquad\text{(d) }\frac15
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{16}+\frac{y^2}{25}=1 \Rightarrow a^2=25,\quad b^2=16.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac{16}{25}}=\frac35.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{The eccentricity of the ellipse } 5x^2+9y^2=1\text{ is}
\displaystyle \text{(a) }\frac23\qquad\text{(b) }\frac34\qquad \text{(c) }\frac45\qquad\text{(d) }\frac12
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{1/5}+\frac{y^2}{1/9}=1 \Rightarrow a^2=\frac15,\quad b^2=\frac19.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac59}=\frac23.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{The eccentricity of the ellipse } 4x^2+9y^2=36\text{ is}
\displaystyle \text{(a) }\frac1{2\sqrt3}\qquad\text{(b) }\frac1{\sqrt3} \qquad\text{(c) }\frac{\sqrt5}{3}\qquad\text{(d) }\frac{\sqrt5}{6}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1 \Rightarrow a^2=9,\quad b^2=4.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac49}=\frac{\sqrt5}{3}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{Equation of the hyperbola whose vertices are } (\pm3,0)
\displaystyle \text{and foci at }(\pm5,0),\text{ is}
\displaystyle \text{(a) }16x^2-9y^2=144\qquad \text{(b) }9x^2-16y^2=144
\displaystyle \text{(c) }25x^2-9y^2=225\qquad \text{(d) }9x^2-25y^2=81
\displaystyle \text{Answer:}
\displaystyle a=3,\quad c=5.
\displaystyle c^2=a^2+b^2\Rightarrow b^2=25-9=16.
\displaystyle \therefore\frac{x^2}{9}-\frac{y^2}{16}=1.
\displaystyle \therefore16x^2-9y^2=144.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{If }e_1\text{ and }e_2\text{ are respectively the eccentricities}
\displaystyle \text{of the ellipse }\frac{x^2}{18}+\frac{y^2}{4}=1 \text{ and the hyperbola }\frac{x^2}{9}-\frac{y^2}{4}=1,\text{ then}
\displaystyle \text{the relation between }e_1\text{ and }e_2\text{ is}
\displaystyle \text{(a) }3e_1^2+e_2^2=2\qquad \text{(b) }e_1^2+2e_2^2=3
\displaystyle \text{(c) }2e_1^2+e_2^2=3\qquad \text{(d) }e_1^2+3e_2^2=2
\displaystyle \text{Answer:}
\displaystyle e_1^2=1-\frac4{18}=\frac79,\qquad e_2^2=1+\frac49=\frac{13}{9}.
\displaystyle 2e_1^2+e_2^2 =2\left(\frac79\right)+\frac{13}{9}=3.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{The distance between the directrices of the hyperbola}
\displaystyle x=8\sec\theta,\ y=8\tan\theta,\text{ is}
\displaystyle \text{(a) }8\sqrt2\qquad\text{(b) }16\sqrt2\qquad \text{(c) }4\sqrt2\qquad\text{(d) }6\sqrt2
\displaystyle \text{Answer:}
\displaystyle x^2-y^2=64\Rightarrow\frac{x^2}{64}-\frac{y^2}{64}=1.
\displaystyle a=8,\quad b=8,\quad e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt2.
\displaystyle \text{Directrices are }x=\pm\frac{a}{e} =\pm\frac8{\sqrt2}=\pm4\sqrt2.
\displaystyle \therefore\text{Distance between the directrices}=8\sqrt2.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{The equation of the conic with focus at }(1,-1),
\displaystyle \text{directrix along }x-y+1=0\text{ and eccentricity }\sqrt2\text{ is}
\displaystyle \text{(a) }xy=1\qquad\text{(b) }2xy+4x-4y-1=0
\displaystyle \text{(c) }x^2-y^2=1\qquad \text{(d) }2xy-4x+4y+1=0
\displaystyle \text{Answer:}
\displaystyle (x-1)^2+(y+1)^2 =2\left(\frac{x-y+1}{\sqrt2}\right)^2.
\displaystyle (x-1)^2+(y+1)^2=(x-y+1)^2.
\displaystyle x^2+y^2-2x+2y+2 =x^2+y^2+1-2xy+2x-2y.
\displaystyle \therefore2xy-4x+4y+1=0.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 51: }\text{The eccentricity of the conic } 9x^2-16y^2=144\text{ is}
\displaystyle \text{(a) }\frac54\qquad\text{(b) }\frac43\qquad \text{(c) }\frac45\qquad\text{(d) }\sqrt7
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{16}-\frac{y^2}{9}=1 \Rightarrow a^2=16,\quad b^2=9.
\displaystyle e=\sqrt{1+\frac{b^2}{a^2}} =\sqrt{1+\frac9{16}}=\frac54.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{A point moves in a plane so that its distances }PA \text{ and }PB
\displaystyle \text{from two fixed points }A\text{ and }B\text{ satisfy the relation } PA-PB=k\ (k\ne0),
\displaystyle \text{then the locus of }P\text{ is}
\displaystyle \text{(a) a hyperbola}\qquad \text{(b) a branch of the hyperbola}
\displaystyle \text{(c) a parabola}\qquad \text{(d) an ellipse}
\displaystyle \text{Answer:}
\displaystyle PA-PB=k,\quad k\ne0.
\displaystyle \text{The difference of the distances of }P\text{ from two fixed points}
\displaystyle \text{is constant. Hence, the locus of }P\text{ is a hyperbola.}
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{The difference of the focal distances of any point}
\displaystyle \text{on the hyperbola is equal to}
\displaystyle \text{(a) length of the conjugate axis}\qquad \text{(b) eccentricity}
\displaystyle \text{(c) length of the transverse axis}\qquad \text{(d) latus-rectum}
\displaystyle \text{Answer:}
\displaystyle \text{For a hyperbola, the difference of the focal distances}
\displaystyle \text{of any point is }2a.
\displaystyle \text{Also, the length of the transverse axis}=2a.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{The foci of the hyperbola } 9x^2-16y^2=144\text{ are}
\displaystyle \text{(a) }(\pm4,0)\qquad\text{(b) }(0,\pm4)\qquad \text{(c) }(\pm5,0)\qquad\text{(d) }(0,\pm5)
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{16}-\frac{y^2}{9}=1 \Rightarrow a^2=16,\quad b^2=9.
\displaystyle c^2=a^2+b^2=16+9=25\Rightarrow c=5.
\displaystyle \therefore\text{Foci are }(\pm5,0).
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{The distance between the foci of a hyperbola is }16
\displaystyle \text{and its eccentricity is }\sqrt2,\text{ then equation of the hyperbola is}
\displaystyle \text{(a) }x^2+y^2=32\qquad\text{(b) }x^2-y^2=16
\displaystyle \text{(c) }x^2+y^2=16\qquad\text{(d) }x^2-y^2=32
\displaystyle \text{Answer:}
\displaystyle 2ae=16\Rightarrow ae=8.
\displaystyle e=\sqrt2\Rightarrow a=\frac8{\sqrt2}=4\sqrt2 \Rightarrow a^2=32.
\displaystyle b^2=a^2(e^2-1)=32(2-1)=32.
\displaystyle \frac{x^2}{32}-\frac{y^2}{32}=1 \Rightarrow x^2-y^2=32.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{If }e_1\text{ is the eccentricity of the conic } 9x^2+4y^2=36
\displaystyle \text{and }e_2\text{ is the eccentricity of the conic } 9x^2-4y^2=36,\text{ then}
\displaystyle \text{(a) }e_1^2-e_2^2=2\qquad \text{(b) }2<e_2^2-e_1^2<3
\displaystyle \text{(c) }e_2^2-e_1^2=2\qquad \text{(d) }e_2^2-e_1^2>3
\displaystyle \text{Answer:}
\displaystyle 9x^2+4y^2=36\Rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1.
\displaystyle e_1^2=1-\frac49=\frac59.
\displaystyle 9x^2-4y^2=36\Rightarrow \frac{x^2}{4}-\frac{y^2}{9}=1.
\displaystyle e_2^2=1+\frac94=\frac{13}{4}.
\displaystyle e_2^2-e_1^2=\frac{13}{4}-\frac59 =\frac{117-20}{36}=\frac{97}{36}.
\displaystyle 2<\frac{97}{36}<3.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{If the eccentricity of the hyperbola } x^2-y^2\sec^2\alpha=5
\displaystyle \text{is }\sqrt3\text{ times the eccentricity of the ellipse } x^2\sec^2\alpha+y^2=25,\text{ then }\alpha=
\displaystyle \text{(a) }\frac{\pi}{6}\qquad\text{(b) }\frac{\pi}{4}\qquad \text{(c) }\frac{\pi}{3}\qquad\text{(d) }\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle x^2-y^2\sec^2\alpha=5 \Rightarrow\frac{x^2}{5}-\frac{y^2}{5\cos^2\alpha}=1.
\displaystyle e_H^2=1+\cos^2\alpha.
\displaystyle x^2\sec^2\alpha+y^2=25 \Rightarrow\frac{x^2}{25\cos^2\alpha}+\frac{y^2}{25}=1.
\displaystyle e_E^2=1-\cos^2\alpha=\sin^2\alpha.
\displaystyle e_H=\sqrt3\,e_E \Rightarrow1+\cos^2\alpha=3\sin^2\alpha.
\displaystyle 1+\cos^2\alpha=3(1-\cos^2\alpha) \Rightarrow\cos^2\alpha=\frac12.
\displaystyle \therefore\alpha=\frac{\pi}{4}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 58: }\text{The equation of the hyperbola whose foci are } (6,4)\text{ and }(-4,4)
\displaystyle \text{and eccentricity }2,\text{ is}
\displaystyle \text{(a) }\frac{(x-1)^2}{25/4}-\frac{(y-4)^2}{75/4}=1
\displaystyle \text{(b) }\frac{(x+1)^2}{25/4}-\frac{(y+4)^2}{75/4}=1
\displaystyle \text{(c) }\frac{(x-1)^2}{75/4}-\frac{(y-4)^2}{25/4}=1 \qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Centre}=\left(\frac{6-4}{2},\frac{4+4}{2}\right)=(1,4).
\displaystyle 2c=10\Rightarrow c=5,\qquad e=\frac ca=2 \Rightarrow a=\frac52.
\displaystyle a^2=\frac{25}{4},\qquad b^2=c^2-a^2=25-\frac{25}{4}=\frac{75}{4}.
\displaystyle \therefore\frac{(x-1)^2}{25/4} -\frac{(y-4)^2}{75/4}=1.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{The length of the straight line }x-3y=1
\displaystyle \text{intercepted by the hyperbola }x^2-4y^2=1\text{ is}
\displaystyle \text{(a) }\frac6{\sqrt5}\qquad \text{(b) }3\sqrt{\frac25}\qquad \text{(c) }6\sqrt{\frac25}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x-3y=1\Rightarrow x=1+3y.
\displaystyle (1+3y)^2-4y^2=1 \Rightarrow5y^2+6y=0.
\displaystyle y=0,-\frac65\Rightarrow P=(1,0),\quad Q=\left(-\frac{13}{5},-\frac65\right).
\displaystyle PQ=\sqrt{\left(1+\frac{13}{5}\right)^2+ \left(\frac65\right)^2}.
\displaystyle PQ=\frac65\sqrt{10}=6\sqrt{\frac25}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{The latus-rectum of the hyperbola } 16x^2-9y^2=144\text{ is}
\displaystyle \text{(a) }\frac{16}{3}\qquad \text{(b) }\frac{32}{3}\qquad \text{(c) }\frac83\qquad\text{(d) }\frac43
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{9}-\frac{y^2}{16}=1 \Rightarrow a=3,\quad b^2=16.
\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a} =\frac{2(16)}3=\frac{32}{3}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 61: }\text{The foci of the hyperbola } 2x^2-3y^2=5\text{ are}
\displaystyle \text{(a) }\left(\pm\frac5{\sqrt6},0\right)\qquad \text{(b) }\left(\pm\frac56,0\right)
\displaystyle \text{(c) }\left(\pm\frac{\sqrt5}{6},0\right)\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{5/2}-\frac{y^2}{5/3}=1 \Rightarrow a^2=\frac52,\quad b^2=\frac53.
\displaystyle c^2=a^2+b^2=\frac52+\frac53=\frac{25}{6}.
\displaystyle \therefore c=\frac5{\sqrt6}.
\displaystyle \therefore\text{Foci are }\left(\pm\frac5{\sqrt6},0\right).
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{The eccentricity of the hyperbola } x=\frac a2\left(t+\frac1t\right),
\displaystyle y=\frac a2\left(t-\frac1t\right)\text{ is}
\displaystyle \text{(a) }\sqrt2\qquad\text{(b) }\sqrt3\qquad \text{(c) }2\sqrt3\qquad\text{(d) }3\sqrt2
\displaystyle \text{Answer:}
\displaystyle x+y=at,\qquad x-y=\frac a t.
\displaystyle (x+y)(x-y)=a^2\Rightarrow x^2-y^2=a^2.
\displaystyle \therefore\frac{x^2}{a^2}-\frac{y^2}{a^2}=1.
\displaystyle e=\sqrt{1+\frac{a^2}{a^2}}=\sqrt2.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 63: }\text{The equation of the hyperbola whose centre is }(6,2),
\displaystyle \text{one focus is }(4,2)\text{ and eccentricity }2\text{ is}
\displaystyle \text{(a) }3(x-6)^2-(y-2)^2=3\qquad \text{(b) }(x-6)^2-3(y-2)^2=1
\displaystyle \text{(c) }(x-6)^2-2(y-2)^2=1\qquad \text{(d) }2(x-6)^2-(y-2)^2=1
\displaystyle \text{Answer:}
\displaystyle \text{Centre }=(6,2),\quad\text{focus }=(4,2) \Rightarrow c=2.
\displaystyle e=\frac ca=2\Rightarrow a=\frac c2=1.
\displaystyle c^2=a^2+b^2\Rightarrow b^2=4-1=3.
\displaystyle \therefore\frac{(x-6)^2}{1}-\frac{(y-2)^2}{3}=1.
\displaystyle \therefore3(x-6)^2-(y-2)^2=3.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 64: }\text{The locus of the point of intersection of the lines}
\displaystyle \sqrt3x-y-4\sqrt3\lambda=0\text{ and } \sqrt3\lambda x+\lambda y-4\sqrt3=0\text{ is a}
\displaystyle \text{hyperbola of eccentricity}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad \text{(c) }3\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \sqrt3x-y=4\sqrt3\lambda.
\displaystyle \lambda(\sqrt3x+y)=4\sqrt3.
\displaystyle \text{Eliminating }\lambda,\quad (\sqrt3x-y)(\sqrt3x+y)=48.
\displaystyle 3x^2-y^2=48 \Rightarrow\frac{x^2}{16}-\frac{y^2}{48}=1.
\displaystyle a^2=16,\quad b^2=48.
\displaystyle e=\sqrt{1+\frac{b^2}{a^2}} =\sqrt{1+\frac{48}{16}}=2.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{Write the axis of symmetry of the parabola }y^2=x.
\displaystyle \text{Answer:}
\displaystyle y^2=x\Rightarrow y^2=4ax,\text{ where }4a=1.
\displaystyle \therefore\text{The axis of symmetry is }y=0,\text{ i.e., the x-axis.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the distance between the vertex and focus of}
\displaystyle \text{the parabola }y^2+6y+2x+5=0.
\displaystyle \text{Answer:}
\displaystyle y^2+6y+2x+5=0.
\displaystyle (y+3)^2=-2(x-2).
\displaystyle \text{Comparing with }(y-k)^2=-4a(x-h),\text{ we get }4a=2.
\displaystyle \therefore a=\frac12.
\displaystyle \text{Hence, the distance between the vertex and focus is }\frac12\text{ unit.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Write the equation of the directrix of the parabola}
\displaystyle x^2-4x-8y+12=0.
\displaystyle \text{Answer:}
\displaystyle x^2-4x-8y+12=0.
\displaystyle (x-2)^2=8(y-1).
\displaystyle \text{Comparing with }(x-h)^2=4a(y-k),\text{ we get }a=2.
\displaystyle \therefore\text{Directrix is }y=k-a=1-2=-1.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Write the equation of the parabola with focus }(0,0)
\displaystyle \text{and directrix }x+y-4=0.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y)\text{ be any point on the parabola.}
\displaystyle \sqrt{x^2+y^2}=\frac{|x+y-4|}{\sqrt2}.
\displaystyle 2(x^2+y^2)=(x+y-4)^2.
\displaystyle \therefore x^2+y^2-2xy+8x+8y-16=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Write the length of the chord of parabola }y^2=4ax
\displaystyle \text{which passes through the vertex and is inclined to the axis at }\frac{\pi}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{The chord through the vertex has slope }\tan\frac{\pi}{4}=1.
\displaystyle \therefore y=x.
\displaystyle x^2=4ax\Rightarrow x=0\text{ or }x=4a.
\displaystyle \text{Thus, the other end of the chord is }(4a,4a).
\displaystyle \therefore\text{Chord length}=\sqrt{(4a)^2+(4a)^2}=4\sqrt2\,a.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }b\text{ and }c\text{ are lengths of the segments of any}
\displaystyle \text{focal chord of the parabola }y^2=4ax,\text{ then write the length of its}
\displaystyle \text{latus-rectum.}
\displaystyle \text{Answer:}
\displaystyle \text{For a focal chord, }\frac1b+\frac1c=\frac1a.
\displaystyle \therefore a=\frac{bc}{b+c}.
\displaystyle \text{Length of latus-rectum}=4a=\frac{4bc}{b+c}.
\displaystyle \\

\displaystyle \textbf{Question 7: }PSQ\text{ is a focal chord of the parabola }y^2=8x.
\displaystyle \text{If }SP=6,\text{ then write }SQ.
\displaystyle \text{Answer:}
\displaystyle y^2=8x\Rightarrow4a=8\Rightarrow a=2.
\displaystyle \text{For a focal chord, }\frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a}.
\displaystyle \frac16+\frac{1}{SQ}=\frac12\Rightarrow\frac{1}{SQ}=\frac13.
\displaystyle \therefore SQ=3.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write the coordinates of the vertex of the parabola}
\displaystyle \text{whose focus is at }(-2,1)\text{ and directrix is the line }x+y-3=0.
\displaystyle \text{Answer:}
\displaystyle \text{The perpendicular from }(-2,1)\text{ to }x+y-3=0\text{ meets it at }(0,3).
\displaystyle \text{The vertex is the midpoint of }(-2,1)\text{ and }(0,3).
\displaystyle \therefore V=\left(\frac{-2+0}{2},\frac{1+3}{2}\right)=(-1,2).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If the coordinates of the vertex and focus of a}
\displaystyle \text{parabola are }(-1,1)\text{ and }(2,3)\text{ respectively, then write the}
\displaystyle \text{equation of its directrix.}
\displaystyle \text{Answer:}
\displaystyle V=(-1,1),\quad S=(2,3)\Rightarrow\overrightarrow{VS}=(3,2).
\displaystyle \text{If }D\text{ is the foot on the directrix, }V\text{ is the midpoint of }SD.
\displaystyle \therefore D=2V-S=(-4,-1).
\displaystyle \text{The directrix passes through }(-4,-1)\text{ and has normal }(3,2).
\displaystyle 3(x+4)+2(y+1)=0.
\displaystyle \therefore\text{Directrix: }3x+2y+14=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If the parabola }y^2=4ax\text{ passes through the point}
\displaystyle (3,2),\text{ then find the length of its latusrectum.}
\displaystyle \text{Answer:}
\displaystyle (3,2)\text{ lies on }y^2=4ax.
\displaystyle 2^2=4a(3)\Rightarrow4=12a\Rightarrow a=\frac13.
\displaystyle \therefore\text{Length of latusrectum}=4a=\frac43.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Write the equation of the parabola whose vertex is}
\displaystyle \text{at }(-3,0)\text{ and the directrix is }x+5=0.
\displaystyle \text{Answer:}
\displaystyle \text{Vertex}=(-3,0),\qquad\text{directrix }x=-5.
\displaystyle a=(-3)-(-5)=2.
\displaystyle \text{Using }(y-k)^2=4a(x-h),
\displaystyle y^2=8(x+3).
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Write the eccentricity of the ellipse}
\displaystyle 9x^2+5y^2-18x-2y-16=0.
\displaystyle \text{Answer:}
\displaystyle 9(x-1)^2+5\left(y-\frac15\right)^2=\frac{126}{5}.
\displaystyle \therefore\frac{(x-1)^2}{14/5} +\frac{\left(y-\frac15\right)^2}{126/25}=1.
\displaystyle a^2=\frac{126}{25},\qquad b^2=\frac{14}{5}.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac59}=\frac23.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Write the centre and eccentricity of the ellipse}
\displaystyle 3x^2+4y^2-6x+8y-5=0.
\displaystyle \text{Answer:}
\displaystyle 3(x^2-2x)+4(y^2+2y)-5=0.
\displaystyle 3(x-1)^2+4(y+1)^2=12.
\displaystyle \therefore\frac{(x-1)^2}{4}+\frac{(y+1)^2}{3}=1.
\displaystyle \text{Centre}=(1,-1),\quad a^2=4,\quad b^2=3.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac34}=\frac12.
\displaystyle \\

\displaystyle \textbf{Question 14: }PSQ\text{ is a focal chord of the ellipse }4x^2+9y^2=36
\displaystyle \text{such that }SP=4.\text{ If }S'\text{ is the another focus, write the value of }S'Q.
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1\Rightarrow a=3,\quad b^2=4.
\displaystyle \text{For a focal chord, }\frac1{SP}+\frac1{SQ}=\frac{2a}{b^2}.
\displaystyle \frac14+\frac1{SQ}=\frac64=\frac32.
\displaystyle \therefore\frac1{SQ}=\frac54\Rightarrow SQ=\frac45.
\displaystyle SQ+S'Q=2a=6.
\displaystyle \therefore S'Q=6-\frac45=\frac{26}{5}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Write the eccentricity of an ellipse whose}
\displaystyle \text{latus-rectum is one half of the minor axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a},\quad \text{minor axis}=2b.
\displaystyle \frac{2b^2}{a}=\frac12(2b)\Rightarrow \frac{2b^2}{a}=b\Rightarrow b=\frac a2.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac14}=\frac{\sqrt3}{2}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If the distance between the foci of an ellipse is}
\displaystyle \text{equal to the length of the latus-rectum, write its eccentricity.}
\displaystyle \text{Answer:}
\displaystyle \text{Distance between foci}=2ae,\quad \text{latus-rectum}=\frac{2b^2}{a}.
\displaystyle 2ae=\frac{2b^2}{a}\Rightarrow e=\frac{b^2}{a^2}.
\displaystyle \text{But }\frac{b^2}{a^2}=1-e^2.
\displaystyle \therefore e=1-e^2\Rightarrow e^2+e-1=0.
\displaystyle e=\frac{-1+\sqrt5}{2}=\frac{\sqrt5-1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }S\text{ and }S'\text{ are two foci of the ellipse}
\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ and }B\text{ is an end of the minor axis such that}
\displaystyle \triangle BSS'\text{ is equilateral, then write the eccentricity of the ellipse.}
\displaystyle \text{Answer:}
\displaystyle S=(ae,0),\quad S'=(-ae,0),\quad B=(0,b).
\displaystyle SS'=2ae,\qquad BS=\sqrt{a^2e^2+b^2}=a.
\displaystyle \triangle BSS'\text{ is equilateral}\Rightarrow SS'=BS.
\displaystyle 2ae=a\Rightarrow e=\frac12.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If the minor axis of an ellipse subtends an}
\displaystyle \text{equilateral triangle with vertex at one end of major axis, then write}
\displaystyle \text{the eccentricity of the ellipse.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=(a,0),\quad B=(0,b),\quad B'=(0,-b).
\displaystyle BB'=2b,\qquad AB=\sqrt{a^2+b^2}.
\displaystyle \triangle ABB'\text{ is equilateral}\Rightarrow AB=BB'.
\displaystyle \sqrt{a^2+b^2}=2b\Rightarrow a^2=3b^2.
\displaystyle \therefore\frac{b^2}{a^2}=\frac13.
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{\frac23}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If a latus-rectum of an ellipse subtends a right angle}
\displaystyle \text{at the centre of the ellipse, then write the eccentricity of the ellipse.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the ends of the latus-rectum be } L\left(ae,\frac{b^2}{a}\right),\ L'\left(ae,-\frac{b^2}{a}\right).
\displaystyle \angle LOL'=90^\circ\Rightarrow (ae)^2-\left(\frac{b^2}{a}\right)^2=0.
\displaystyle a^2e=\;b^2.
\displaystyle \therefore e=\frac{b^2}{a^2}=1-e^2.
\displaystyle e^2+e-1=0\Rightarrow e=\frac{\sqrt5-1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If the lengths of semi-major and semi-minor axes}
\displaystyle \text{of an ellipse are }2\text{ and }\sqrt3\text{ and their corresponding equations}
\displaystyle \text{are }y-5=0\text{ and }x+3=0,\text{ then write the equation of the ellipse.}
\displaystyle \text{Answer:}
\displaystyle \text{Centre}=(-3,5),\quad a=2,\quad b=\sqrt3.
\displaystyle \text{The major axis is }y=5,\text{ so it is parallel to the }x\text{-axis}.
\displaystyle \frac{(x+3)^2}{4}+\frac{(y-5)^2}{3}=1.
\displaystyle 3(x+3)^2+4(y-5)^2=12.
\displaystyle 3x^2+4y^2+18x-40y+115=0.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Write the eccentricity of the hyperbola } 9x^2-16y^2=144.
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{16}-\frac{y^2}{9}=1 \Rightarrow a^2=16,\quad b^2=9.
\displaystyle e=\sqrt{1+\frac{b^2}{a^2}} =\sqrt{1+\frac9{16}}=\frac54.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Write the eccentricity of the hyperbola whose}
\displaystyle \text{latus-rectum is half of its transverse axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Latus-rectum}=\frac{2b^2}{a},\quad \text{transverse axis}=2a.
\displaystyle \frac{2b^2}{a}=\frac12(2a) \Rightarrow b^2=\frac{a^2}{2}.
\displaystyle e=\sqrt{1+\frac{b^2}{a^2}} =\sqrt{1+\frac12}=\sqrt{\frac32}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Write the coordinates of the foci of the hyperbola}
\displaystyle 9x^2-16y^2=144.
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{16}-\frac{y^2}{9}=1 \Rightarrow a^2=16,\quad b^2=9.
\displaystyle c^2=a^2+b^2=16+9=25\Rightarrow c=5.
\displaystyle \therefore\text{Foci are }(5,0)\text{ and }(-5,0).
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Write the equation of the hyperbola of eccentricity }\sqrt2,
\displaystyle \text{if it is known that the distance between its foci is }16.
\displaystyle \text{Answer:}
\displaystyle 2ae=16\Rightarrow ae=8.
\displaystyle e=\sqrt2\Rightarrow a=\frac8{\sqrt2}=4\sqrt2 \Rightarrow a^2=32.
\displaystyle b^2=a^2(e^2-1)=32(2-1)=32.
\displaystyle \therefore\frac{x^2}{32}-\frac{y^2}{32}=1.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If the foci of the ellipse } \frac{x^2}{16}+\frac{y^2}{b^2}=1\text{ and the hyperbola}
\displaystyle \frac{x^2}{144}-\frac{y^2}{81}=\frac1{25} \text{ coincide, write the value of }b^2.
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{144/25}-\frac{y^2}{81/25}=1.
\displaystyle c^2=\frac{144}{25}+\frac{81}{25}=9.
\displaystyle \text{For the ellipse, }c^2=16-b^2.
\displaystyle 16-b^2=9\Rightarrow b^2=7.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Write the length of the latus-rectum of the hyperbola}
\displaystyle 16x^2-9y^2=144.
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{9}-\frac{y^2}{16}=1 \Rightarrow a=3,\quad b^2=16.
\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a} =\frac{2(16)}3=\frac{32}{3}.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If the latus-rectum through one focus of a hyperbola}
\displaystyle \text{subtends a right angle at the farther vertex, then write its eccentricity.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the hyperbola be }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.
\displaystyle \text{The focus is }S(ae,0)\text{ and the farther vertex is }A'(-a,0).
\displaystyle \text{Ends of the latus-rectum are } L\left(ae,\frac{b^2}{a}\right),\ L'\left(ae,-\frac{b^2}{a}\right).
\displaystyle \angle LA'L'=90^\circ\Rightarrow a^2(e+1)^2-\frac{b^4}{a^2}=0.
\displaystyle a(e+1)=\frac{b^2}{a}=a(e^2-1).
\displaystyle e+1=(e-1)(e+1)\Rightarrow e-1=1.
\displaystyle \therefore e=2.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Write the distance between the directrices of the hyperbola}
\displaystyle x=8\sec\theta,\quad y=8\tan\theta.
\displaystyle \text{Answer:}
\displaystyle x^2-y^2=64\Rightarrow\frac{x^2}{64}-\frac{y^2}{64}=1.
\displaystyle a=8,\quad b=8,\quad e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt2.
\displaystyle \text{Directrices are }x=\pm\frac{a}{e} =\pm\frac8{\sqrt2}=\pm4\sqrt2.
\displaystyle \therefore\text{Distance between the directrices}=8\sqrt2.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Write the equation of the hyperbola whose vertices}
\displaystyle \text{are }(\pm3,0)\text{ and foci at }(\pm5,0).
\displaystyle \text{Answer:}
\displaystyle \text{Vertices }(\pm a,0)=(\pm3,0)\Rightarrow a=3.
\displaystyle \text{Foci }(\pm ae,0)=(\pm5,0)\Rightarrow ae=5.
\displaystyle b^2=a^2(e^2-1)=(ae)^2-a^2=25-9=16.
\displaystyle \therefore\frac{x^2}{9}-\frac{y^2}{16}=1.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If }e_1\text{ and }e_2\text{ are respectively the eccentricities}
\displaystyle \text{of the ellipse }\frac{x^2}{18}+\frac{y^2}{4}=1 \text{ and the hyperbola }\frac{x^2}{9}-\frac{y^2}{4}=1,
\displaystyle \text{then write the value of }2e_1^2+e_2^2.
\displaystyle \text{Answer:}
\displaystyle e_1^2=1-\frac4{18}=\frac79.
\displaystyle e_2^2=1+\frac49=\frac{13}{9}.
\displaystyle \therefore2e_1^2+e_2^2 =2\left(\frac79\right)+\frac{13}{9}=\frac{27}{9}=3.
\displaystyle \\

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