Two straight lines are perpendicular to each other. One of them touches the parabola $y^{2} = 4 a (x + a)$ and the other touches $y^{2} = 4 b (x + b)$ . Then the locus of point of intersection of two lines is

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x + a = 0

x – a –b = 0

x + b = 0

x + a + b = 0

answer is C.

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Detailed Solution

$\begin{array}{l} y^{2} = 4 a (x + a) \Rightarrow y = m (x + a) + \frac{a}{m} \\ \Rightarrow y = mx + am + \frac{a}{m} \dots . . (1) \\ E q u a t i o n o f \tan g e n t t o t h e p a r a b o l a y^{2} = 4 b (x + b) i s \\ y = \frac{- (x + b)}{m} - b m \\ y = \frac{- x}{m} - \frac{b}{m} - b m . . . . . . . (2) \\ f r o m (1) a n d (2) i s x = - a - b \end{array}$