The equation of the conic with focus $S (3, 0)$ and the directix $x + 3 = 0$ having eccentricity $1$ , is

x^{2} = 6 y

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x^{2} = 12 y

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y^{2} = 6 y

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y^{2} = 12 y

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Solution

The correct option is B $x^{2} = 12 y$
$G i v e n$
$f o c u s S (3, 0)$
$d i r e c t i x : x + 3$
$e = 1$
$sin c e e = 1, h e n c e the given conic is a parabola$
$we are given focus S= (3, 0)$
$h e n c e i t i s p a r a b o l a o f f o r m x^{2} = 4 a y$
$w i t h f o c u s = (a, 0)$
$h e n c e a = 3$
$t h u s t h e g i v e n c o n i c i s : x^{2} = 4 (3) y$
$x^{2} = 12 y$

Suggest Corrections

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