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

The point on the curve $y^2=2(x-3)$ at which the normal is parallel to the line $y-2 x+1=0$ is
  1. $\left(\frac{-1}{2},-2\right)$
  2. $\left(\frac{3}{2}, 2\right)$
  3. $(5,2)$
  4. $(5,-2)$

Solution

$\begin{aligned} & y^2=2(x-3) \\ & \therefore 2 y \frac{d y}{d x}=2 \quad \Rightarrow \frac{d y}{d x}=\frac{1}{y} \\ & \therefore \text { Slope of normal }=-y \text { and as per condition given } \\ & -y=2 \quad \Rightarrow y=-2 \\ & \therefore \quad(-2)^2=2(x-3) \Rightarrow x=5 \Rightarrow \text { point is }(5,-2)\end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 1)

Practice more Applications of Derivatives questions on Aicharya