Class 11: Parabola – Multiple Choice Questions

$\displaystyle \textbf{Question 1. }\text{The coordinates of the focus of the parabola } \\ y^2-x-2y+2=0\text{ are}$
$\displaystyle \text{(a) }\left(\frac{5}{4},1\right)\qquad \text{(b) }\left(\frac{1}{4},0\right)\qquad \text{(c) }(1,1)\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle y^2-x-2y+2=0$
$\displaystyle y^2-2y+1=x-1$
$\displaystyle (y-1)^2=x-1$
$\displaystyle (y-1)^2=4A(x-1)$
$\displaystyle \therefore 4A=1,\quad A=\frac{1}{4}$
$\displaystyle \text{Vertex}=(1,1)$
$\displaystyle \therefore \text{Focus}=\left(1+\frac{1}{4},1\right)=\left(\frac{5}{4},1\right)$
$\displaystyle \therefore \text{Correct option is (a)}$
$\\$

$\displaystyle \textbf{Question 2. }\text{The vertex of the parabola }(y+a)^2=8a(x-a)\text{ is}$
$\displaystyle \text{(a) }(-a,-a)\qquad \text{(b) }(a,-a)\qquad \text{(c) }(-a,a)\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle (y+a)^2=8a(x-a)$
$\displaystyle (y-(-a))^2=8a(x-a)$
$\displaystyle \text{Comparing with }(y-k)^2=4A(x-h)$
$\displaystyle h=a,\quad k=-a$
$\displaystyle \therefore \text{Vertex}=(a,-a)$
$\displaystyle \therefore \text{Correct option is (b)}$
$\\$

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

$\displaystyle \textbf{Question 4. }\text{The equation of the parabola whose vertex is }(a,0)\text{ and the} \\ \text{directrix has the equation }x+y=3a,\text{ is}$
$\displaystyle \text{(a) }x^2+y^2+2xy+6ax+10ay+7a^2=0$
$\displaystyle \text{(b) }x^2-2xy+y^2+6ax+10ay-7a^2=0$
$\displaystyle \text{(c) }x^2-2xy+y^2-6ax+10ay-7a^2=0$
$\displaystyle \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Directrix is }x+y-3a=0$
$\displaystyle \text{Foot of perpendicular from }(a,0)\text{ to }x+y-3a=0\text{ is }(2a,a)$
$\displaystyle \text{Since vertex is midpoint of focus and foot,}$
$\displaystyle (a,0)=\left(\frac{x_1+2a}{2},\frac{y_1+a}{2}\right)$
$\displaystyle \therefore \text{Focus}=(0,-a)$
$\displaystyle \text{For any point }(x,y)\text{ on the parabola,}$
$\displaystyle \sqrt{x^2+(y+a)^2}=\frac{|x+y-3a|}{\sqrt{2}}$
$\displaystyle 2\{x^2+(y+a)^2\}=(x+y-3a)^2$
$\displaystyle 2x^2+2y^2+4ay+2a^2=x^2+y^2+9a^2+2xy-6ax-6ay$
$\displaystyle x^2-2xy+y^2+6ax+10ay-7a^2=0$
$\displaystyle \therefore \text{Correct option is (b)}$
$\\$

$\displaystyle \textbf{Question 5. }\text{The parametric equations of a parabola are }x=t^2+1,\\ y=2t+1.\text{ The cartesian equation of its directrix is}$
$\displaystyle \text{(a) }x=0\qquad \text{(b) }x+1=0\qquad \text{(c) }y=0\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle x=t^2+1,\quad y=2t+1$
$\displaystyle y-1=2t$
$\displaystyle t=\frac{y-1}{2}$
$\displaystyle x=\left(\frac{y-1}{2}\right)^2+1$
$\displaystyle (y-1)^2=4(x-1)$
$\displaystyle \text{Comparing with }(y-k)^2=4A(x-h)$
$\displaystyle h=1,\quad k=1,\quad A=1$
$\displaystyle \therefore \text{Directrix is }x=h-A=1-1=0$
$\displaystyle \therefore \text{Correct option is (a)}$
$\\$

$\displaystyle \textbf{Question 6. }\text{If the coordinates of the vertex and the focus of a parabola are }(-1,1)\text{ and }(2,3) \\ \text{ respectively, then the equation of its directrix is}$
$\displaystyle \text{(a) }3x+2y+14=0\qquad \text{(b) }3x+2y-25=0\qquad \text{(c) }2x-3y+10=0\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Vertex}=(-1,1),\quad \text{Focus}=(2,3)$
$\displaystyle \text{Axis has direction }(2+1,3-1)=(3,2)$
$\displaystyle \text{Directrix is perpendicular to the axis}$
$\displaystyle \therefore \text{Directrix is of the form }3x+2y+c=0$
$\displaystyle \text{Let foot of perpendicular from focus to directrix be }(x_1,y_1)$
$\displaystyle \text{Vertex is midpoint of focus and foot}$
$\displaystyle (-1,1)=\left(\frac{2+x_1}{2},\frac{3+y_1}{2}\right)$
$\displaystyle x_1=-4,\quad y_1=-1$
$\displaystyle \therefore (-4,-1)\text{ lies on directrix}$
$\displaystyle 3(-4)+2(-1)+c=0$
$\displaystyle -12-2+c=0$
$\displaystyle c=14$
$\displaystyle \therefore \text{Directrix is }3x+2y+14=0$
$\displaystyle \therefore \text{Correct option is (a)}$
$\\$

$\displaystyle \textbf{Question 7. }\text{The locus of the points of trisection of the double ordinates of a parabola is a}$
$\displaystyle \text{(a) pair of lines}\qquad \text{(b) circle}\qquad \text{(c) parabola}\qquad \text{(d) straight line}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the parabola be }y^2=4ax$
$\displaystyle \text{Ends of a double ordinate are }(x,y)\text{ and }(x,-y)$
$\displaystyle \text{Points of trisection have ordinates }\frac{y}{3}\text{ and }-\frac{y}{3}$
$\displaystyle \text{Let }Y=\frac{y}{3}$
$\displaystyle y=3Y$
$\displaystyle \therefore (3Y)^2=4ax$
$\displaystyle 9Y^2=4ax$
$\displaystyle Y^2=\frac{4a}{9}x$
$\displaystyle \therefore \text{The locus is a parabola}$
$\displaystyle \therefore \text{Correct option is (c)}$
$\\$

$\displaystyle \textbf{Question 8. }\text{The equation of the directrix of the parabola whose vertex and focus are} \\ (1,4)\text{ and }(2,6)\text{ respectively is}$
$\displaystyle \text{(a) }x+2y=4\qquad \text{(b) }x-y=3\qquad \text{(c) }2x+y=5\qquad \text{(d) }x+3y=8$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Vertex}=(1,4),\quad \text{Focus}=(2,6)$
$\displaystyle \text{Axis has direction }(2-1,6-4)=(1,2)$
$\displaystyle \text{Directrix is perpendicular to the axis}$
$\displaystyle \therefore \text{Directrix is of the form }x+2y+c=0$
$\displaystyle \text{Let foot of perpendicular from focus to directrix be }(x_1,y_1)$
$\displaystyle \text{Vertex is midpoint of focus and foot}$
$\displaystyle (1,4)=\left(\frac{2+x_1}{2},\frac{6+y_1}{2}\right)$
$\displaystyle x_1=0,\quad y_1=2$
$\displaystyle \therefore (0,2)\text{ lies on directrix}$
$\displaystyle 0+2(2)+c=0$
$\displaystyle c=-4$
$\displaystyle \therefore \text{Directrix is }x+2y-4=0$
$\displaystyle \therefore x+2y=4$
$\displaystyle \therefore \text{Correct option is (a)}$
$\\$

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

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

$\displaystyle \textbf{Question 11. }\text{The equation of the parabola with focus }(0,0)\text{ and directrix } \\ x+y=4\text{ is}$
$\displaystyle \text{(a) }x^2+y^2-2xy+8x+8y-16=0$
$\displaystyle \text{(b) }x^2+y^2-2xy+8x+8y=0$
$\displaystyle \text{(c) }x^2+y^2+8x+8y-16=0$
$\displaystyle \text{(d) }x^2-y^2+8x+8y-16=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For any point }(x,y)\text{ on the parabola,}$
$\displaystyle \text{distance from }(x,y)\text{ to focus}=\text{distance from }(x,y)\text{ to directrix}$
$\displaystyle \sqrt{x^2+y^2}=\frac{|x+y-4|}{\sqrt{2}}$
$\displaystyle 2(x^2+y^2)=(x+y-4)^2$
$\displaystyle 2x^2+2y^2=x^2+y^2+16+2xy-8x-8y$
$\displaystyle x^2+y^2-2xy+8x+8y-16=0$
$\displaystyle \therefore \text{Correct option is (a)}$
$\\$

$\displaystyle \textbf{Question 12. }\text{The line }2x-y+4=0\text{ cuts the parabola }y^2=8x\text{ in }P\text{ and }Q. \\ \text{ The mid-point of }PQ\text{ is}$
$\displaystyle \text{(a) }(1,2)\qquad \text{(b) }(1,-2)\qquad \text{(c) }(-1,2)\qquad \text{(d) }(-1,-2)$
$\displaystyle \text{Answer:}$
$\displaystyle 2x-y+4=0$
$\displaystyle y=2x+4$
$\displaystyle y^2=8x$
$\displaystyle (2x+4)^2=8x$
$\displaystyle 4x^2+16x+16=8x$
$\displaystyle 4x^2+8x+16=0$
$\displaystyle x^2+2x+4=0$
$\displaystyle \text{If }x_1,x_2\text{ are the abscissae of }P\text{ and }Q,\text{ then}$
$\displaystyle x_1+x_2=-2$
$\displaystyle \therefore \text{Mid-point abscissa}=\frac{x_1+x_2}{2}=-1$
$\displaystyle y_1+y_2=2(x_1+x_2)+8=2(-2)+8=4$
$\displaystyle \therefore \text{Mid-point ordinate}=\frac{y_1+y_2}{2}=2$
$\displaystyle \therefore \text{Mid-point of }PQ=(-1,2)$
$\displaystyle \therefore \text{Correct option is (c)}$
$\\$

$\displaystyle \textbf{Question 13. }\text{In the parabola }y^2=4ax,\text{ the length of the chord passing through} \\ \text{the vertex and inclined to the axis at }\frac{\pi}{4}\text{ is}$
$\displaystyle \text{(a) }4\sqrt{2}a\qquad \text{(b) }2\sqrt{2}a\qquad \text{(c) }\sqrt{2}a\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Axis of }y^2=4ax\text{ is the }x\text{-axis}$
$\displaystyle \text{Chord through vertex and inclined at }\frac{\pi}{4}\text{ to the axis is }y=x$
$\displaystyle \text{Putting }y=x\text{ in }y^2=4ax$
$\displaystyle x^2=4ax$
$\displaystyle x(x-4a)=0$
$\displaystyle x=0\text{ or }x=4a$
$\displaystyle \therefore \text{Points are }(0,0)\text{ and }(4a,4a)$
$\displaystyle \therefore \text{Length of chord}=\sqrt{(4a)^2+(4a)^2}$
$\displaystyle =4\sqrt{2}a$
$\displaystyle \therefore \text{Correct option is (a)}$
$\\$

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

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

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

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

$\displaystyle \textbf{Question 18. }\text{The length of the latus-rectum of the parabola } \\ 4y^2+2x-20y+17=0\text{ is}$
$\displaystyle \text{(a) }3\qquad \text{(b) }6\qquad \text{(c) }\frac{1}{2}\qquad \text{(d) }9$
$\displaystyle \text{Answer:}$
$\displaystyle 4y^2+2x-20y+17=0$
$\displaystyle 4(y^2-5y)+2x+17=0$
$\displaystyle 4\left(y^2-5y+\frac{25}{4}\right)+2x+17=25$
$\displaystyle 4\left(y-\frac{5}{2}\right)^2+2x-8=0$
$\displaystyle 4\left(y-\frac{5}{2}\right)^2=-2x+8$
$\displaystyle \left(y-\frac{5}{2}\right)^2=-\frac{1}{2}(x-4)$
$\displaystyle \text{Comparing with }(y-k)^2=4A(x-h)$
$\displaystyle 4A=-\frac{1}{2}$
$\displaystyle \therefore \text{Length of latus-rectum}=|4A|=\frac{1}{2}$
$\displaystyle \therefore \text{Correct option is (c)}$
$\\$

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

$\displaystyle \textbf{Question 20. }\text{The focus of the parabola }y=2x^2+x\text{ is}$
$\displaystyle \text{(a) }(0,0)\qquad \text{(b) }\left(\frac{1}{2},\frac{1}{4}\right)\qquad \text{(c) }\left(-\frac{1}{4},0\right)\qquad \text{(d) }\left(-\frac{1}{4},\frac{1}{8}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle y=2x^2+x$
$\displaystyle y=2\left(x^2+\frac{x}{2}\right)$
$\displaystyle y=2\left(x+\frac{1}{4}\right)^2-\frac{1}{8}$
$\displaystyle y+\frac{1}{8}=2\left(x+\frac{1}{4}\right)^2$
$\displaystyle \left(x+\frac{1}{4}\right)^2=\frac{1}{2}\left(y+\frac{1}{8}\right)$
$\displaystyle \text{Comparing with }(x-h)^2=4A(y-k)$
$\displaystyle h=-\frac{1}{4},\quad k=-\frac{1}{8},\quad 4A=\frac{1}{2}$
$\displaystyle A=\frac{1}{8}$
$\displaystyle \therefore \text{Focus}=(h,k+A)=\left(-\frac{1}{4},-\frac{1}{8}+\frac{1}{8}\right)$
$\displaystyle =\left(-\frac{1}{4},0\right)$
$\displaystyle \therefore \text{Correct option is (c)}$
$\\$

$\displaystyle \textbf{Question 21. }\text{Which of the following points lie on the parabola }x^2=4ay\text{?}$
$\displaystyle \text{(a) }x=at^2,\ y=2at\qquad \text{(b) }x=2at,\ y=at^2$
$\displaystyle \text{(c) }x=2at^2,\ y=at\qquad \text{(d) }x=2at,\ y=at^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }x^2=4ay,\text{ the standard parametric point is}$
$\displaystyle x=2at,\quad y=at^2$
$\displaystyle \text{Checking: }x^2=(2at)^2=4a^2t^2$
$\displaystyle 4ay=4a(at^2)=4a^2t^2$
$\displaystyle \therefore x^2=4ay$
$\displaystyle \therefore \text{Correct option is (b) and (d)}$
$\\$

$\displaystyle \textbf{Question 22. }\text{The equation of the parabola whose focus is }(1,-1)\text{ and the} \\ \text{directrix is }x+y+7=0\text{ is}$
$\displaystyle \text{(a) }x^2+y^2-2xy-18x-10y=0$
$\displaystyle \text{(b) }x^2-18x-10y-45=0$
$\displaystyle \text{(c) }x^2+y^2-18x-10y-45=0$
$\displaystyle \text{(d) }x^2+y^2-2xy-18x-10y-45=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For any point }(x,y)\text{ on the parabola,}$
$\displaystyle \sqrt{(x-1)^2+(y+1)^2}=\frac{|x+y+7|}{\sqrt{2}}$
$\displaystyle 2\{(x-1)^2+(y+1)^2\}=(x+y+7)^2$
$\displaystyle 2(x^2-2x+1+y^2+2y+1)=x^2+y^2+49+2xy+14x+14y$
$\displaystyle 2x^2+2y^2-4x+4y+4=x^2+y^2+2xy+14x+14y+49$
$\displaystyle x^2+y^2-2xy-18x-10y-45=0$
$\displaystyle \therefore \text{Correct option is (d)}$
$\\$