If the normal at point P(t) to the parabola y | vedclass

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(D) For the parabola $y^2 = 16x$,we have $4a = 16$,so $a = 4$.Point $Q(36, -24)$ is on the parabola,so $(36, -24) = (at_2^2, 2at_2)$.$2at_2 = -24$ $\R

Vedclass · Accepted Answer

$20$

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(D) For the parabola $y^2 = 16x$,we have $4a = 16$,so $a = 4$.Point $Q(36, -24)$ is on the parabola,so $(36, -24) = (at_2^2, 2at_2)$.$2at_2 = -24$ $\Rightarrow 2(4)t_2 = -24$ $\Rightarrow t_2 = -3$.The relation between the parameters of the normal at $t_1$ meeting the parabola at $t_2$ is $t_2 = -t_1 - \frac{2}{t_1}$.Substituting $t_2 = -3$,we get $-3 = -t_1 - \frac{2}{t_1} \Rightarrow t_1^2 - 3t_1 + 2 = 0$.Solving for $t_1$,we get $(t_1 - 1)(t_1 - 2) = 0$,so $t_1 = 1$ or $t_1 = 2$.The focal distance of point $P(at_1^2, 2at_1)$ is $a(1 + t_1^2)$.For $t_1 = 1$,focal distance $= 4(1 + 1^2) = 8$.For $t_1 = 2$,focal distance $= 4(1 + 2^2) = 20$.The maximum focal distance is $20$.

If the normal at point $P(t)$ to the parabola $y^2 = 16x$ meets it again at point $Q(36, -24)$,then the maximum possible focal distance of point $P$ is-

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