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 at $Q$. Then $OP\cdot OQ=$

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

Answer 1

The correct option is D $4a^2$

Here $O$ is the vertex,
Equation of tangent at $(a{t}^2,2at)$ is $yt=x+a{t}^2\cdots \left(1\right)$
Distance
$OP=\frac{|0-0-a{t}^2|}{\sqrt{1+t^2}}\Rightarrow OP=\frac{|a{t}^2|}{\sqrt{1+t^2}}$

Equation of perpendicular to line $\left(1\right)$from $O(0,0)$ is
$y+tx=0\cdots \left(2\right)$
Now, intersection of $y^2=4ax$ and line $\left(2\right)$ is
$Q(\frac{4a}{t^2},\frac{-4a}{t})$
Distance
$OQ=\sqrt{{\left(\frac{4a}{t^2}\right)}^2+{\left(\frac{-4a}{t}\right)}^2}\Rightarrow OQ=|4a|\sqrt{\frac{1}{t^4}+\frac{1}{t^2}}\Rightarrow OQ=|4a|\frac{\sqrt{1+t^2}}{t^2}$
Now,
$OP\cdot OQ=\frac{|a{t}^2|}{\sqrt{1+t^2}}\times |4a|\frac{\sqrt{1+t^2}}{t^2}\therefore OP\cdot OQ=4a^2$