A point P is such that the straight line drawn through it perpendicular to its polar with respect to the parabola $y^{2} = 4 a x$ touches the parabola $x^{2} = 4 b y$ . Prove that its locus is the straight line
$2 a x + b y + 4 a^{2} = 0$

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Solution

Let the point P be $(h, k)$
Its polar w.r.t $y^{2} = 4 a x$ will be $y k = 2 a x + 2 a h$
The line passing through $(h, k)$ perpendicular to the line $y k = 2 a x + 2 a h$ would be $y = \frac{- k}{2 a} x + k + \frac{h k}{2 a}$
i.e. $2 a y + k x = 2 a k + h k$
A tangent to the parabola $x^{2} = 4 b y$ is $t x = y + b t^{2}$
Comparing the two equations, we have
$\frac{k}{t} = \frac{2 a}{- 1} = \frac{2 a k + h k}{b t^{2}}$
$\Rightarrow t = \frac{- k}{2 a}$
Also, $- 2 a b t^{2} = 2 a k + h k$
$∴ - 2 a b \times \frac{k^{2}}{4 a^{2}} = 2 a k + h k$
$∴ - b k^{2} = 2 a (2 a k + h k)$
i.e. $b k + 4 a^{2} + 2 a h = 0$
$b y + 4 a^{2} + 2 a x = 0$ is the required locus.

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