Question

$P$ 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:

• A. $6$
• B. $2$
• C. $4$
• D. $8$

Answer 1

The correct option is D $8$

Let $P(t^2,2t)\&Q(t_1^2,2t_1)y^2=4x\Rightarrow \frac{dy}{dx}=\frac{2}{y}$
Slope of tangent at $P=\frac{2}{2t}=\frac{1}{t}$
Slope of normal at $P=-t$
Equation of normal at point $P(t^2,2t),$$y-2t=-t(x-t^2)$
It passes through $Q\left(t_1\right)$
$2t_1-2t=-t(t_1^2-t^2)\Rightarrow t_1=-t-\frac{2}{t}$
Squaring both the sides, we get,
$t_1^2=t^2+\frac{4}{t^2}+4.......\left(1\right)$
We know that, $AM\ge GM$$\Rightarrow \frac{t^2+\frac{4}{t^2}}{2}\ge \sqrt{t^2\times \frac{4}{t^2}}{t}^2+\frac{4}{t^2}\ge 4......\left(2\right)$
From eq$\left(1\right)$ and $\left(2\right)$, we get,
$t_1^2\ge 8\Rightarrow$ Mini. value of $t_1^2=8$