Question

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 $O N \cdot O M=$

• A. $4 a^{2}$
• B. $3 a^{2}$
• C. $2 a^{2}$
• D. $a^{2}$

Answer 1

Answer: (A)

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

$y t=x+ a t^{2}$ ...(i)
$\because N M$ is normal to the tangent Eq. (i) and passes through origin ' $O^{\prime}$ so equation of line $N M$ is
$y=-t x$ ...(ii)
So, points $N \equiv\left(-\frac{a t^{2}}{1+t^{2}}, \frac{a t^{3}}{1+t^{2}}\right)$
and $M \equiv\left(\frac{4 a}{t^{2}},-\frac{4 a}{t}\right)$
So, $O N \cdot O M =\sqrt{\frac{a^{2} t^{4}}{\left(1+t^{2}\right)^{2}}+\frac{a^{2} t^{6}}{\left(1+t^{2}\right)^{2}}} \times \sqrt{\frac{16 a^{2}}{t^{4}}+\frac{16 a^{2}}{t^{2}}}$
$=\frac{4 a^{2} t^{2}}{t\left(1+t^{2}\right)} \sqrt{1+t^{2}} \sqrt{\frac{1}{t^{2}}+1}=4 a^{2}$