If $\left( \alpha ,\beta \right)$ is the point on $y{}^{2}$ = 6x, that is closest to $\left( 3,\dfrac{3}{2} \right)$ then find 2 $\left( \alpha +\beta \right)$ ?
(A) 6
(B) 9
(C) 7
(D) 5

Hint: in this question, we have to find the value of 2 $\left( \alpha +\beta \right)$. To solve this, first, we take the equation of normal at the given points then we pass the normal from the given closest points. Also a parabola is given which is satisfied by $\left( \alpha ,\beta \right)$. By taking all the equations and solving them, we find the value of $\alpha $ and $\beta $. Now we put these values in 2 $\left( \alpha +\beta \right)$and by solving it, we get our answer.

Formula Used:
Equation of normal at $\left( {{x}_{1}},{{y}_{1}} \right)$is
$y-{{y}_{1}}=m(x-{{x}_{1}})$

Complete step by step Solution:
Given that $\left( \alpha ,\beta \right)$ is the point on $y{}^{2}$ = 6x
We have to find the value of 2 $\left( \alpha +\beta \right)$
We know equation of normal at $\left( {{x}_{1}},{{y}_{1}} \right)$is
$y-{{y}_{1}}=m(x-{{x}_{1}})$
Where m is the slope. Then
Equation of normal at $\left( \alpha ,\beta \right)$ is
$y-\beta =\left( \dfrac{-\beta }{2a} \right)(x-\alpha )$
Solving the above equation, we get
$3y-3\beta =-\beta (x-\alpha )$------------------ (1)
For shortest distance normal will pass through $\left( 3,\dfrac{3}{2} \right)$
So, equation (1) becomes
$\dfrac{9}{2}-3\beta =-3\beta -3\alpha $
$\alpha \beta =\dfrac{9}{2}$----------------- (2)
Now from $y{}^{2}$= 6x
If $\left( \alpha ,\beta \right)$ satisfies the above equation, we get
${{\beta }^{2}}=6\alpha $
$\alpha $$\alpha =\dfrac{{{\beta }^{2}}}{6}$--------------------- (3)
From equations (2) and (3), we get
$\dfrac{{{\beta }^{2}}\times \beta }{6}=\dfrac{9}{2}$
By solving the above equation, we get
${{\beta }^{3}}=27$
Taking cube roots on both sides, we get
$\beta =3$
$\alpha =\dfrac{3}{2},\beta =3$
Now we find the value of 2 $\left( \alpha +\beta \right)$
For this we put the value of $\alpha $ and $\beta $ in the above equation, we get
2 $\left( \alpha +\beta \right)$ = 2$\left( \dfrac{3}{2}+3 \right)$
Which is equal to 2$\left( \dfrac{9}{2} \right)$ = 9
Hence the value of 2 $\left( \alpha +\beta \right)$ = 9

Hence, the correct option is B.

Note: Students made mistakes in putting the values from one equation to another. There are many methods to solve this question but in all, we have to put the values from one equation to another. Be careful while putting these values.