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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

If the vertex of the parabola is the point $(-3,0)$ and the directrix is the line $x+5=0$,then its equation is

$\begin{array}{1 1}y^2=8(x+3)\\x^2=8(y+3)\\y^2=-8(x+3)\\y^2=8(x+5)\end{array} $

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A)
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  • General equation of a parabola whose vertex is $(h,k)$ is $(y-k)^2=4a(x-h)$
Given the vertex of the parabola is $(-3,0)$
The equation of directrix is $x+5=0$
Equation of the parabola is $(y-0)^2=4a(x-(3))$
(ie) $y^2=4a(x+3)$
$x=-a$ is the equation of the directrix .
Given directrix is $x+5$
(ie) $x+5-3=0$
(ie) $x+2=0$
$\therefore a=2$
$\therefore$ Equation of the required parabola is $y^2=8(x+3)$
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