Class 11: Parabola – Very Short Answer Questions

$\displaystyle \textbf{Question 1. }\text{Write the axis of symmetry of the parabola }y^2=x.$
$\displaystyle \text{Answer:}$
$\displaystyle y^2=x\text{ is of the form }y^2=4ax$
$\displaystyle \text{The axis of symmetry of }y^2=4ax\text{ is the }x\text{-axis}$
$\displaystyle \therefore \text{Axis of symmetry is }y=0$
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$\displaystyle \textbf{Question 2. }\text{Write the distance between the vertex and focus of the parabola} \\ y^2+6y+2x+5=0.$
$\displaystyle \text{Answer:}$
$\displaystyle y^2+6y+2x+5=0$
$\displaystyle \Rightarrow y^2+6y=-2x-5$
$\displaystyle \Rightarrow y^2+6y+9=-2x-5+9$
$\displaystyle \Rightarrow (y+3)^2=-2(x-2)$
$\displaystyle \text{Comparing with }(y-k)^2=4a(x-h)$
$\displaystyle 4a=-2$
$\displaystyle \therefore a=-\frac{1}{2}$
$\displaystyle \text{Distance between vertex and focus }=|a|=\frac{1}{2}$
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$\displaystyle \textbf{Question 3. }\text{Write the equation of the directrix of the parabola} \\ x^2-4x-8y+12=0.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-4x-8y+12=0$
$\displaystyle \Rightarrow x^2-4x=-12+8y$
$\displaystyle \Rightarrow x^2-4x+4=8y-8$
$\displaystyle \Rightarrow (x-2)^2=8(y-1)$
$\displaystyle \text{Comparing with }(x-h)^2=4a(y-k)$
$\displaystyle 4a=8\Rightarrow a=2$
$\displaystyle \text{Directrix is }y=k-a$
$\displaystyle \therefore y=1-2=-1$
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$\displaystyle \textbf{Question 4. }\text{Write the equation of the parabola with focus }(0,0)\text{ and directrix} \\ x+y-4=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the parabola}$
$\displaystyle \text{Distance from }P(x,y)\text{ to focus }(0,0)=\sqrt{x^2+y^2}$
$\displaystyle \text{Distance from }P(x,y)\text{ to directrix }x+y-4=0\text{ is }\frac{|x+y-4|}{\sqrt{2}}$
$\displaystyle \therefore \sqrt{x^2+y^2}=\frac{|x+y-4|}{\sqrt{2}}$
$\displaystyle \Rightarrow 2(x^2+y^2)=(x+y-4)^2$
$\displaystyle \therefore \text{Equation of parabola is }2(x^2+y^2)=(x+y-4)^2$
$\displaystyle 2(x^2+y^2)=(x+y-4)^2$
$\displaystyle 2x^2+2y^2=(x+y)^2-8(x+y)+16$
$\displaystyle 2x^2+2y^2=x^2+2xy+y^2-8x-8y+16$
$\displaystyle x^2+y^2-2xy+8x+8y-16=0$
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$\displaystyle \textbf{Question 5. }\text{Write the length of the chord of the parabola }y^2=4ax\text{ which passes} \\ \text{through the vertex and is inclined to the axis at }\frac{\pi}{4}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The axis of }y^2=4ax\text{ is the }x\text{-axis}$
$\displaystyle \text{The chord passes through the vertex and is inclined at }\frac{\pi}{4}\text{ to the axis}$
$\displaystyle \therefore \text{Equation of chord is }y=x$
$\displaystyle \text{Substituting }y=x\text{ in }y^2=4ax$
$\displaystyle x^2=4ax$
$\displaystyle \Rightarrow x(x-4a)=0$
$\displaystyle \therefore x=0\text{ or }x=4a$
$\displaystyle \text{Points of intersection are }(0,0)\text{ and }(4a,4a)$
$\displaystyle \therefore \text{Length of chord}=\sqrt{(4a-0)^2+(4a-0)^2}$
$\displaystyle =4a\sqrt{2}$
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$\displaystyle \textbf{Question 6. }\text{If }b\text{ and }c\text{ are lengths of the segments of any focal chord of} \\ \text{the parabola }y^2=4ax,\text{ then write the length of its latus-rectum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the parabola }y^2=4ax,\text{ length of latus-rectum}=4a$
$\displaystyle \text{If }b\text{ and }c\text{ are the segments of a focal chord, then }\frac{1}{b}+\frac{1}{c}=\frac{1}{a}$
$\displaystyle \Rightarrow \frac{b+c}{bc}=\frac{1}{a}$
$\displaystyle \Rightarrow a=\frac{bc}{b+c}$
$\displaystyle \therefore \text{Length of latus-rectum}=4a=\frac{4bc}{b+c}$
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$\displaystyle \textbf{Question 7. }PSQ\text{ is a focal chord of the parabola }y^2=8x.\text{ If }SP=6,\text{ then write }SQ.$
$\displaystyle \text{Answer:}$
$\displaystyle y^2=8x\Rightarrow 4a=8\Rightarrow a=2$
$\displaystyle \text{For a focal chord, }\frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a}$
$\displaystyle \therefore \frac{1}{6}+\frac{1}{SQ}=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{1}{SQ}=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$
$\displaystyle \therefore SQ=3$
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$\displaystyle \textbf{Question 8. }\text{Write the coordinates of the vertex of the parabola whose focus is at} \\ (-2,1)\text{ and directrix is the line }x+y-3=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Vertex lies midway between the focus and the foot of perpendicular from the focus to the directrix.}$
$\displaystyle \text{Equation of directrix: }x+y-3=0$
$\displaystyle \text{Foot of perpendicular from }(x_1,y_1)\text{ to }ax+by+c=0\text{ is}$
$\displaystyle \left(\frac{x_1-a(ax_1+by_1+c)}{a^2+b^2},\frac{y_1-b(ax_1+by_1+c)}{a^2+b^2}\right)$
$\displaystyle \text{Here }a=1,\ b=1,\ c=-3,\ (x_1,y_1)=(-2,1)$
$\displaystyle ax_1+by_1+c=-2+1-3=-4$
$\displaystyle \therefore \text{Foot of perpendicular }=\left(\frac{-2-1(-4)}{1^2+1^2},\frac{1-1(-4)}{1^2+1^2}\right)$
$\displaystyle =\left(\frac{2}{2},\frac{5}{2}\right)=\left(1,\frac52\right)$
$\displaystyle \text{Vertex is midpoint of }(-2,1)\text{ and }\left(1,\frac52\right)$
$\displaystyle \therefore \text{Vertex}=\left(\frac{-2+1}{2},\frac{1+\frac52}{2}\right)$
$\displaystyle =\left(-\frac12,\frac74\right)$

$\displaystyle \textbf{Question 9. }\text{If the coordinates of the vertex and focus of a parabola are }(-1,1) \\ \text{ and }(2,3) \text{ respectively, then write the equation of its directrix.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Vertex }V(-1,1)\text{ is the midpoint of focus and foot of perpendicular on directrix}$
$\displaystyle \text{Let the foot of perpendicular be }D(x,y)$
$\displaystyle \therefore \left(\frac{2+x}{2},\frac{3+y}{2}\right)=(-1,1)$
$\displaystyle \Rightarrow x=-4,\ y=-1$
$\displaystyle \therefore D=(-4,-1)$
$\displaystyle \text{Slope of axis }VF=\frac{3-1}{2-(-1)}=\frac{2}{3}$
$\displaystyle \therefore \text{Slope of directrix}=-\frac{3}{2}$
$\displaystyle \text{Equation of directrix through }(-4,-1)\text{ is}$
$\displaystyle y+1=-\frac{3}{2}(x+4)$
$\displaystyle \Rightarrow 3x+2y+14=0$
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$\displaystyle \textbf{Question 10. }\text{If the parabola }y^2=4ax\text{ passes through the point }(3,2),\text{ then} \\ \text{find the length of its latusrectum.}$
$\displaystyle \text{Answer:}$
$\displaystyle y^2=4ax$
$\displaystyle \text{Since it passes through }(3,2)$
$\displaystyle 2^2=4a(3)$
$\displaystyle \Rightarrow 4=12a$
$\displaystyle \Rightarrow a=\frac{1}{3}$
$\displaystyle \therefore \text{Length of latusrectum}=4a=\frac{4}{3}$
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$\displaystyle \textbf{Question 11. }\text{Write the equation of the parabola whose vertex is at }(-3,0)\text{ and} \\ \text{the directrix is }x+5=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Directrix is }x=-5$
$\displaystyle \text{Vertex is }(-3,0)$
$\displaystyle \therefore a=2$
$\displaystyle \text{Parabola opens to the right and has form }(y-k)^2=4a(x-h)$
$\displaystyle \therefore (y-0)^2=4(2)(x+3)$
$\displaystyle \therefore y^2=8(x+3)$
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