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Pair of tangent are drawn to the parabola $y^{2} = - 4 x$ from every point on the line $3 x + y = 2$ then prove that their chord of contact passes through a fixed point.

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Solution

$\begin{matrix} k y = - 4 \frac{(x + h)}{2} k y = - 2 (x + h) 2 y + 2 x + h (2 - 3 y) = 0 x + y = 02 - 3 y = 0 x = \frac{- 2}{3} y = \frac{2}{3} H e n c e, c h o r d o f c o n t a c t p a s s e s t h r o u g h k (\frac{- 2}{3}, \frac{2}{3}) . \end{matrix}$

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