The normal chord at a point 't' on the parabola $y^{2}$ = $4 a x$ subtends a right angle at the vertex . Then $t^{2}$ is equal to

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Solution

The correct option is B 2
let end point of the chord is $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$

since it is normal to parabola so $t_{2} = - t_{1} - \frac{2}{t_{1}}$

equation of normal of parabola $y = - t x + 2 a t + a t^{3}$

slope is $- t_{1}$ which is given acute so $- t_{1} > 0$

or $t_{1}$

now chord make right angel at origin so $\frac{2 a t_{1} - 0}{a t_{1}^{2} - 0} \frac{2 a t_{1} - 0}{a t_{2}^{2} - 0} = - 1$

$t_{1} t_{2} = - 4$

$t_{1} (- t_{1} - \frac{2}{t_{1}}) = - 4$

$t_{1} = - \sqrt{2}$

Hence

$t^{2} = 2$

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