Question

Through the vertex O of the parabola $y^2=4ax,a$ perpendicular is drawn to any tangent meeting it at P and the parabola T Q find the value of OP .OQ?

Answer 1

We have,

Equation of tangent

$y^2=4ax$ at point $(a{t}^2,2at)$

Now

$yt=x+a{t}^2$

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

Then, length of perpendicular from $O(0,0)$ is

$OP=\left|\frac{x-yt+a{t}^2}{\sqrt{1+t^2}}\right|$

$OP=\left|\frac{0x-0t+a{t}^2}{\sqrt{1+t^2}}\right|$

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

Now, Equation of perpendicular to $\left(1\right)$ is,

$tx+y=k$

It passes through $(0,0)$

Now,

$0+0=k$

$\Rightarrow k=0$

Then,

$tx+y=0$

$\Rightarrow y=-tx.....\left(2\right)$

Put in equation of parabola

$y^2=4ax$

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

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

From equation

$y=-tx$

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

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

Using distance formula

$OQ=\sqrt{{(\frac{4a}{t^2}-0)}^2+{(-\frac{4a}{t}-0)}^2}$

$=4a\sqrt{\frac{1}{t^4}+\frac{1}{t^2}}$

$=4a\sqrt{\frac{1+t^2}{t^4}}$

$OP.OQ=\frac{a{t}^2}{\sqrt{1+t^2}}\frac{4a}{t^2}\sqrt{1+t^2}=4a^2$

Hence, this is the answer.