If the point on the curve y ^ 2 = 6x, nearest to the point (3,...

We have,

${y}^{2} = 6x$

Comparing with ${y}^{2} = 4ax$, we get $a = \dfrac{3}{2}$

Let the point on parabola be $P ≡ \left(a{t}^{2}, 2at\right) ≡ \left(\dfrac{3}{2}{t}^{2}, 3t\right)$.

Normal at point $P$ is

$tx + y = 3t + \dfrac{3}{2}{t}^{3}$

It passes through \left(3, \dfrac{3}{2}\right)$, hence</p><p>$⇒ 3t + \dfrac{3}{2} = 3t + \dfrac{3}{2}{t}^{3}$</p><p>$⇒ {t}^{3} = 1 ⇒ t = 1$</p><p>Hence,</p><p>$P ≡ \left(\dfrac{3}{2}, 3\right) = ( \alpha, \beta )$</p><p>Therefore,</p><p>$2 ( \alpha + \beta ) = 2\left(\dfrac{3}{2} + 3\right) = 9