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If a perpendicular drawn through the vertex $O$ of the parabola $y^2=4 a x$ to any of its tangent meets the tangent at $N$ and the parabola at $M$, then $O N \cdot O M=$
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$4 a^2$
Let a point $P\left(a t^2, 2 a t\right)$ over the parabola $y^2=4 a x$. So, equation of tangent at point $P$ is
$\begin{aligned} & \text { So, points } N \equiv\left(-\frac{a t^2}{1+t^2}, \frac{a t^3}{1+t^2}\right) \\ & \text { and } \quad M \equiv\left(\frac{4 a}{t^2},-\frac{4 a}{t}\right)\end{aligned}$
So, $\begin{aligned} O N \cdot O M & =\sqrt{\frac{a^2 t^4}{\left(1+t^2\right)^2}+\frac{a^2 t^6}{\left(1+t^2\right)^2}} \times \sqrt{\frac{16 a^2}{t^4}+\frac{16 a^2}{t^2}} \\ & =\frac{4 a^2 t^2}{t\left(1+t^2\right)} \sqrt{1+t^2} \sqrt{\frac{1}{t^2}+1}=4 a^2 .\end{aligned}$
$\begin{aligned} & \text { So, points } N \equiv\left(-\frac{a t^2}{1+t^2}, \frac{a t^3}{1+t^2}\right) \\ & \text { and } \quad M \equiv\left(\frac{4 a}{t^2},-\frac{4 a}{t}\right)\end{aligned}$
So, $\begin{aligned} O N \cdot O M & =\sqrt{\frac{a^2 t^4}{\left(1+t^2\right)^2}+\frac{a^2 t^6}{\left(1+t^2\right)^2}} \times \sqrt{\frac{16 a^2}{t^4}+\frac{16 a^2}{t^2}} \\ & =\frac{4 a^2 t^2}{t\left(1+t^2\right)} \sqrt{1+t^2} \sqrt{\frac{1}{t^2}+1}=4 a^2 .\end{aligned}$
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