Question

If the point on the curve $y^{2}=6 x$, nearest to the point $\left(3, \frac{3}{2}\right)$ is $(\alpha, \beta)$, then $2(\alpha+\beta)$ is equal to _______.

Answer 1

$P \equiv\left(\frac{3}{2} t^{2}, 3 t\right)$
Normal at point $P$
$t x+y=3 t+\frac{3}{2} t^{3}$
Passes through $\left(3, \frac{3}{2}\right)$
$\Rightarrow 3 t+\frac{3}{2}=3 t+\frac{3}{2} t^{3}$
$P \equiv\left(\frac{3}{2}, 3\right)=(\alpha, \beta)$
$\Rightarrow t^{3}=1 \Rightarrow t=1$
$2(\alpha+\beta)=2\left(\frac{3}{2}+3\right)=9$