Question

Through the vertex O of the parabola $y^2=4ax$, a perpendicular is drawn to any tangent meeting it at P & the parabola at Q. Show that OP$\cdot$OQ$=$constant.

Answer 1

We assume that $O$ is vertex, $P$ is point where perpendicular meets tangent. Equation of tangent $y^2=4ax$ at $\left(a{t}^{[}2\right],2at)$

$yt=2ta{t}^2$

$\Rightarrow x-yt+a{t}^2=0$ ------- $\left(1\right)$

Length of perpendicular from $O(0,0)$ is

$OP=\frac{|O+O+a{t}^2|}{\sqrt{1+t^2}}$

$OP=\frac{|a{t}^2|}{\sqrt{1+t^2}}$

Equation of perpendicular to $,i,$ is

$tx+y=k$

It passes through $(O,O)$

$O+O=k$

$tx+y=O$

$y=-tx$ put in $y^2=4ax$

$t^2{x}^2=4ax$

$x=\frac{4a}{t^2}$

$y=-tx=-t\frac{4a}{t^2}=-\frac{4a}{t}$

Hence $Q(\frac{4a}{t^2},\frac{4a}{t})$

By distance formula $,QB=\sqrt{{(\frac{4a}{t^2}-0)}^2+{(-\frac{4a}{t}-0)}^2}=|4a|\sqrt{\frac{1}{t^4}+\frac{1}{t^2}}$

$=|4a|\sqrt{\frac{1+t_2}{t_4}}.OQ=\left|\frac{4a}{t^2}\right|\sqrt{1+t^2},OP.OQ=4a^2$.