Question

If a perpendicular drawn through the vertex $O$ of the parabola $y^2=4ax$ to any of its tangent meets the tangent at $N$ and the parabola at $M$, then $ON\cdot OM=$

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

Answer 1

Answer: (A)

Let a point $P\left(a{t}^2,2at\right)$ over the parabola $y^2=4ax.$ So, equation of tangent at point $P$ is

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