Question

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

On differentiating with respect to $x$, we get

$2y\frac{dy}{dx}=2$

$\Rightarrow \frac{dy}{dx}=\frac{1}{y}$

Slope of the normal $=\frac{-1}{\left(\frac{dy}{dx}\right)}=-y$

The slope of the given line $=2$

$\therefore y=-2$

From equation (i), $x=5$

$\therefore$ Required point is $(5,-2)$.