Question

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

Answer 1

Answer: 9

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