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f(x) Transforms to f(-x)

If the normal...

Question

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

T^{2} \geq 8

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- 2 < T < 2

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T^{2} < 8

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none of these

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Solution

The correct option is D $T^{2} \geq 8$
Parabola: $y^{2} = 4 a x . . . . . . . . . . . . (1)$

Normal is drawn at $P : (a t^{2}, 2 a t)$ and it cuts curve again at $(a T^{2}, 2 a T)$

We know when normal at $t_{1}$ cut curve again $t_{2}, (t_{2} = - t_{1} - \frac{2}{t_{1}})$

So, $T = - t - \frac{2}{t}$

$=> t^{2} + T t + 2 = 0$ (quadratic)

For real solutions, $D \geq 0$

$=> T^{2} - 4 (2) (1) \geq 0$

$=> T^{2} \geq 8.$

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