P and Q are two distinct points on the parabo | vedclass

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(A) The equation of the normal to the parabola $y^2 = 4ax$ at parameter $t$ is $y = -tx + 2at + at^3$. Here $a = 1$,so the normal at $P(t)$ is $y = -t

Vedclass · Accepted Answer

$8$

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(A) The equation of the normal to the parabola $y^2 = 4ax$ at parameter $t$ is $y = -tx + 2at + at^3$. Here $a = 1$,so the normal at $P(t)$ is $y = -tx + 2t + t^3$.Since this normal passes through $Q(t_1)$,we have $t_1^2 = -t(t_1) + 2t + t^3$,which simplifies to $t_1^2 - t_1^2 = 0$ is not correct,rather the relation is $t_1 = -t - \frac{2}{t}$.Squaring both sides,we get $t_1^2 = (-t - \frac{2}{t})^2 = t^2 + \frac{4}{t^2} + 4$.Using the Arithmetic Mean-Geometric Mean inequality,$t^2 + \frac{4}{t^2} \ge 2\sqrt{t^2 \cdot \frac{4}{t^2}} = 4$.Therefore,the minimum value of $t_1^2 = 4 + 4 = 8$.

$P$ and $Q$ are two distinct points on the parabola,$y^2 = 4x$,with parameters $t$ and $t_1$ respectively. If the normal at $P$ passes through $Q$,then the minimum value of $t_1^2$ is

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