If parabola $y^{2} = 4 a x$ passes through the point $(9, - 12)$ then find the length of latus rectum and coordinates of focus.

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Solution

$∵ y^{2} = 4 a x p a s s e s t h r o u g h (9, - 12)$
$(- 12)^{2} = 4 a (9)$
$a = \frac{144}{36} \Rightarrow 4$
$\Rightarrow f o c u s o f p a r a b o l a b e (a, 0)$
$∴ f o c u s = (4, 0)$
Length of latus Rectum $= 4 a$
$= 4 \times 4$
$= 16 U n i t s$

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