Prove that the locus of the poles of chords of a parabola which subtend a right angle at a fixed point (h, k) is
$a x^{2} - h y^{2} + (4 a^{2} + 2 a h) x - 2 a k y + a (h^{2} + k^{2}) = 0 .$

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Solution

For $y^{2} = 4 a x$ let the pole be $T (x_{1}, y_{1})$ and the point of contact of tangents be $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$

Equation of chord of contact with respect to $T$ is

$y y_{1} = 2 a (x + x_{1}) 2 a x - y y_{1} + 2 a x_{1} . . . . . . (i)$

Equation of chord joining $P Q$ is

$(t_{1} + t_{2}) y = 2 x + 2 a t_{1} t_{2} 2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0..... (i i)$

Now $(i)$ and $(i i)$ are the equation of same line

$\frac{2}{2 a} = \frac{- (t_{1} + t_{2})}{- y_{1}} = \frac{2 a t_{1} t_{2}}{2 a x_{1}} \Rightarrow t_{1} + t_{2} = \frac{y_{1}}{a}, t_{1} t_{2} = \frac{x_{1}}{a} . . . . . (i i i)$

Given $\frac{k - 2 a t_{1}}{h - a t_{1}^{2}} \times \frac{k - 2 a t_{2}}{h - a t_{2}^{2}} = - 1$

$k^{2} - 2 a k (t_{1} + t_{2}) + 4 a^{2} t_{1} t_{2} = - (h^{2} - a h (t_{1}^{2} + t_{2}^{2}) + a^{2} t_{1}^{2} t_{2}^{2}) k^{2} - 2 a k (t_{1} + t_{2}) + 4 a^{2} t_{1} t_{2} = - {h^{2} - a h ({(t_{1} + t_{2})}^{2} - 2 t_{1} t_{2}) + a^{2} t_{1}^{2} t_{2}^{2}}$

Substituting $(i i i)$

$k^{2} - 2 a k (\frac{y_{1}}{a}) + 4 a^{2} (\frac{x_{1}}{a}) = - {h^{2} - a h ({(\frac{y_{1}}{a})}^{2} - 2 (\frac{x_{1}}{a})) + a^{2} {(\frac{x_{1}}{a})}^{2}} k^{2} - 2 a k (\frac{y_{1}}{a}) + 4 a^{2} (\frac{x_{1}}{a}) = - {h^{2} - a h (\frac{y_{1}^{2} - 2 a x_{1}}{a^{2}}) + a^{2} {(\frac{x_{1}}{a})}^{2}} a k^{2} - 2 a k y_{1} + 4 a^{2} x_{1} = - (a h^{2} - h y_{1}^{2} + 2 h a x_{1} + a x_{1}^{2}) a x_{1}^{2} - h y_{1}^{2} + (4 a^{2} + 2 a h) x_{1} - 2 a k y_{1} + a (k^{2} + h^{2}) = 0$

Generalising the equation

$a x^{2} - h y^{2} + (4 a^{2} + 2 a h) x - 2 a k y + a (k^{2} + h^{2}) = 0$

Hence proved

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