Question

If $(\alpha ,\beta )$ is the point on $y^2=6x$, that is closest to $\left(3,\frac{3}{2}\right)$ then find $2(\alpha +\beta )$

• A. $6$
• B. $9$
• C. $7$
• D. $5$

Answer 1

The correct option is B

$9$

Explanation for the correct option:

Step 1. Draw the figure according to question:

From the figure, we can see

Equation of normal at $(\alpha ,\beta )$ is

$y-\beta =\frac{-\beta }{2a}(x-\alpha )$

$\Rightarrow$$3y-3\beta =-\beta (x-\alpha )$

For the shortest distance normal will pass through $\left(3,\frac{3}{2}\right)$

$\therefore$$\frac{9}{2}–3\beta =-3\beta -\beta \alpha$

$\Rightarrow$ $\alpha \beta =\frac{9}{2}$ ⋯(1)

From $y^2=6x$, we get

${\beta }^2=6\alpha$ ⋯(2)

Step 2. From equation(1) and (2) we get $\alpha =\frac{3}{2},\beta =3$

$\begin{array}{rcl}\therefore 2(\alpha +\beta )& =& 2\left(\frac{3}{2}+3\right)\\ & =& 2\left(\frac{9}{2}\right)\\ & =& \mathbf{9}\end{array}$

Hence, Option ‘B’ is Correct.