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. $\left(-\frac{1}{2},-2\right)$
• C. $(5,-2)$
• D. $\left(\frac{3}{2},2\right)$

Answer 1

The correct option is C

$(5,-2)$

The explanation for the correct answer.

Solve for the value of the point.

Given: $y^2=2(x–3)$………………..$\left(1\right)$

Differentiate with respect to $x$,

$2y\left(\frac{dy}{dx}\right)=2$ $\left[\frac{d{a}^n}{dx}=n{a}^{n-1}\frac{da}{dx}\right]$

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

Since, the slope of the normal is given by, $-\frac{1}{{\displaystyle \frac{dy}{dx}}}=–y$…………………$\left(2\right)$

Normal is parallel to the line, $y-2x+1=0$

It can be written as, $y=2x-1$

We know that for the standard equation of line $y=mx+c$, slope is $m$.

Thus slope of the normal is $2$.

From equation $\left(2\right)$,

$y=-2$

Putting the value of $y$ in equation $\left(1\right)$,

$$\begin{gathered} {(-2)}^2=2(x–3) \\ \Rightarrow 4=2x-6 \\ \Rightarrow 2x=10 \\ \Rightarrow x=5 \end{gathered}$$

Therefore, the required point is $(5,-2)$.

Hence, option (C) is the correct answer.