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Double Ordinate of Hyperbola

If the normal...

Question

If the normal to the parabola $y^{2} = 4 a x$ at the point $P (a t^{2}, 2 a t)$ cuts the parabola again at $Q (a T^{2}, 2 a T)$ , then

- 2 \leq T \leq 2

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T \in (- \infty, - 8) \cup (8, \infty)

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T^{2} < 8

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T^{2} \geq 8

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Solution

The correct option is D $T^{2} \geq 8$
$y + t x = 2 a t + a t^{3}$
( $e q^{n}$ of normal at $a t^{2}, 2 a t$ )
$(a T^{2}, 2 a T)$ lies on the line
$\Rightarrow 2 a T + t a T^{2} = 2 a t + a t^{3}$
On simplifying, we get
$T = - t - \frac{2}{t}$
$T^{2} = {(t + \frac{2}{t})}^{2}$
The minimum value of the expression $(t + \frac{2}{t})$ is $- 2 \sqrt{2}$ at $t = - \sqrt{2}$
$∴ T^{2} ⩾ (2 \sqrt{2})^{2}$
$∴ T^{2} ⩾ 8$

Hence, option D.

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