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)

$-6$

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