The locus of point of intersection of tangents drawn at the ends of chord of $y^{2} = 4 ax$ which subtends a right angle at vertex is

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x + a = 0

x + 2a = 0

x + 4a = 0

y + a = 0

answer is C.

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Detailed Solution

$L e t A (a {t_{1}}^{2}, 2 a t_{1}) a n d B (a {t_{2}}^{2}, 2 a t_{2}) b e t h e e n d s o f t h e c h o r d o f t h e g i v e n p a r a b o l a y^{2} = 4 a x v e r t e x (P) = (0, 0) s l o p e o f P A i s \frac{2}{t_{1}} s l o p e o f P B i s \frac{2}{t_{2}} \sin c e, P A i s p e r p e n d i c u l a r t o P B . T h e r e f o r e, t h e p r o d u c t o f t h e i r s l o p e s i s - 1 \Rightarrow t_{1} t_{2} = - 4$

Let Q $(x_{1}, y_{1}) b e l o c u s o f p o i n t o f i n t e r s e c t i o n o f \tan g e n t s$

$(x_{1}, y_{1}) = ({at}_{1} t_{2}, a (t_{1} + t_{2})) \Rightarrow x_{1} = {at}_{1} t_{2} \Rightarrow x_{1} = - 4 a l o c u s i s x + 4 a = 0$
$\Rightarrow x = - 4 a \Rightarrow x + 4 a = 0$