Question

Find the equation of the normal line to the curve of the equation x 2 y + xy2 = 6 at the point (2, 1).

Answer 1

$$x^2y+xy^2=6$$

Differentiate both sides with respect to $x$

$$\dfrac{d}{dx}(x^2y+xy^2)=\dfrac{d}{dx}(6)$$

Use the Chain Rule

$$2xy+x^2 \dfrac{dy}{dx}+y^2+2xy\dfrac{dy}{dx}=0$$

Solve for $\dfrac{dy}{dx}$

$$\dfrac{dy}{dx}=-\dfrac{2xy+y^2}{x^2+2xy}$$

Find the slope of the tangent line to the curve at the point $(2, 1)$

$${slope}_1=m_1=-\dfrac{2(2)(1)+(1)^2}{(2)^2+2(2)(1)}=-\dfrac{5}{8}$$

Find the slope of the normal line to the curve at the point $(2, 1)$

$${slope}_2=m_2=-\dfrac{1}{m_1}=\dfrac{8}{5}$$

The equation of the normal line in point-slope form

$$y-1=\dfrac{8}{5}(x-2)$$

The equation of the normal line  to the curve of the equation $x ^2 y + xy^2 = 6$ at the point $(2, 1)$ in slope-intercept form

$$y=\dfrac{8}{5}x-\dfrac{11}{5}$$