Question

The slope of the normal to the curve y³ − xy − 8 = 0 at the point (0, 2) is equal to _________________.

Answer 1

The equation of given curve is

y³ − xy − 8 = 0

Differentiating both sides with respect to x, we get

$3y^2\frac{dy}{dx}-\left(x\frac{dy}{dx}+y\right)-0=0$

$\Rightarrow \left(3y^2-x\right)\frac{dy}{dx}=y$

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

Now, slope of tangent at (0, 2) = ${\left(\frac{dy}{dx}\right)}_{\left(0,2\right)}=\frac{2}{3\times {\left(2\right)}^2-0}=\frac{2}{12}=\frac{1}{6}$ .....(1)

∴ Slope of the normal at (0, 2)

$=-\frac{1}{{\left({\displaystyle \frac{dy}{dx}}\right)}_{\left(0,2\right)}}$

$=-\frac{1}{\left({\displaystyle \frac{1}{6}}\right)}$ [From (1)]

$=-6$

Thus, the slope of normal to the curve y³ − xy − 8 = 0 at the point (0, 2) is −6.


The slope of the normal to the curve y³ − xy − 8 = 0 at the point (0, 2) is equal to ___−6___.