Question

The normal at point $\left(1,1\right)$ of the curve $y^2=x^3$ is parallel to the line

• A. $3x-y-2=0$
• B. $2x+3y-7=0$
• C. $2x-3y+1=0$
• D. $2y+3x-1=0$

Answer 1

The correct option is B

$2x+3y-7=0$

Find the line parallel to the given curve

Given, $y^2=x^3$

Differentiating with respect to $x$

$2y\frac{dy}{dx}=3x^2$

At point $\left(1,1\right)$

$2\left(1\right)\frac{dy}{dx}=3{\left(1\right)}^2$

$\frac{dy}{dx}=m_t=\frac{3}{2}$

$m_t{m}_n=-1$, where $m_t,m_n$ are slopes of tangent and normal respectively

$m_n=\frac{-2}{3}$

The slope of the line $2x+3y-7=0$ is given as $m=\frac{-2}{3}$

So, as the slopes are equal the line $2x+3y-7=0$ is parallel to the normal.

Hence, $2x+3y-7=0$ is parallel to the normal, so, option $\left(B\right)$ is the correct answer.