Q. The point of the curve $y^{2} = 2(x - 3)$ at which the normal is parallel to line$\ y - 2x + 1$=0
A) $(5,\ 2)$
B) $\left( - \frac{1}{2},\ - 2 \right)$
C) $(5,\ - 2)$
D) $\left( \frac{3}{2},\ 2 \right)$
Explanation
(c)
Given $y^{2} = 2(x - 3)$ …(i)
$\Rightarrow \ \ \ \ 2y\frac{dy}{dx} = 2\ \ \ \ \Rightarrow \ \ \frac{dy}{dx} = \frac{1}{y}$
Slope of the normal$= \frac{- 1}{(dy/dx)} = - y$
Slope of the given line=2
$\therefore\ \ y = - 2$
From Eq. (i), $x = 5$
$\therefore$ Required point is (5, $- 2$)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.