Question

Let the tangent at point $P(8,4)$ to the curve $y^3=x^2$ meets the curve again at $Q$ and $PQ=a\sqrt{10}.$ Then the value of $a$ is

Answer 1

Given curve : $y^3=x^2$
Let point $Q=(t^3,t^2)$
For slope of tangent at $P:3y^2\frac{dy}{dx}=2x$
So, slope of tangent at $P(8,4)=$ slope of $PQ$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(8,4)}=\frac{t^2-4}{t^3-8}$
$\Rightarrow \frac{1}{3}=\frac{t^2-4}{t^3-8}$
$\Rightarrow \frac{1}{3}=\frac{(t-2)(t+2)}{(t-2)(t^2+2t+4)}$
$\Rightarrow t^2-t-2=0$
$\Rightarrow t=-1,2;(t\ne 2)$
So, $Q\equiv (-1,1)$
$\Rightarrow PQ=\sqrt{81+9}=3\sqrt{10}$
$\therefore a=3$