If a normal chord a point $^{'} t^{'}$ on the parabola $y^{2} = 4 a x$ subtends a right angle at vertex, then prove that $t = \pm \sqrt{2}$

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Solution

Given,

$y^{2} = 4 a x$

Normal at point "t" is given by,

$y = - t x + 2 a t + a t^{3}$

Therefore joint equation is given by,

$y^{2} = 4 a x (\frac{y + t x}{2 a t + a t^{3}})$

$∴ 4 t x^{2} - (2 t + t^{3}) y^{2} + 4 x y = 0$

substituting coeffieients of $x^{2}, y^{2}$ to zero, we get,

$2 t - t^{3} = 0$

$2 t = t^{3}$

$t^{2} = 2$

$∴ t = \pm \sqrt{2}$

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If a normal chord a point $^{'} t^{'}$ on the parabola $y^{2} = 4 a x$ subtends a right angle at vertex, then prove that $t = \pm \sqrt{2}$