Question

The slope of the normal to the curve $y^3-xy-8=0$ at the point $(0,2)$ is equal to

• A. -3
• B. -6
• C. 3
• D. 6
• E. 8

Answer 1

Answer: (B)

Given curve is $y^3-xy-8=0$
On differentiating w.r.t. $x$, we get
$3y^2\frac{dy}{dx}-x\frac{dy}{dx}-y-0=0$
$\Rightarrow \frac{dy}{dx}\left(3y^2-x\right)=y$
$\frac{dy}{dx}=\frac{y}{\left(3y^2-x\right)}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(0,2)}=\frac{2}{3(2{)}^2-0}=\frac{1}{6}$
$\therefore$ The slope of the normal $=-\frac{1}{dy/dx}=-6$