Question

The point of the curve $y^2=2(x-3)$ at which the normal is parallel to the line y-2+1=0 is

• A. (5,2)
• B. $(-\frac{1}{2},-2)$
• C. (5, –2)
• D. $(\frac{3}{2},2)$

Answer 1

The correct option is C (5, –2)

Given $y^2=2(x-3)......\left(i\right)$
Differentiate w.r.t. x, $2y.\frac{dy}{dx}=2\Rightarrow \frac{dy}{dx}=\frac{1}{y}$
Slope of the normal =$\left(\frac{-1}{\left(\frac{dy}{dx}\right)}\right)=-y$
Slope of the given line =2
$\therefore y=-2$
From equation (i), x =5
$\therefore$ Required point is (5,-2).