Finding the normal to a curve

Question

Find the equation of the normal to the curve with equation $4x^2+xy^2-3y^3=56$ at the point $(-5,2)$.

I know that the normal to a curve is $$-\frac{1}{f'(x)}$$ And when I differentiate the curve implicitly I get $$-\frac{8x-y^2}{6y^2}$$

Substituting that into the equation for a normal you get a positive reciprocal $6y^2/(8x-y^2)$ But apparently this is wrong, I'm given the points $(-5,2)$ how are these useful?

Answer 1

I believe your implicit differentiation is wrong. Given $4x^2+xy^2-3y^3=56$, we can implicitly differentiate to find $\frac{dy}{dx}$:

$$ 4x^2+xy^2-3y^3=56 \\ 8x+\color{blue}{\left(y^2+2xy\frac{dy}{dx}\right)} - 9y^2\frac{dy}{dx}=0 \\ $$

(The blue part is arrived at through the product rule.) After simplifying this, we can plug in the point $(-5,2)$ to find the slope of the tangent line at that point. Like you said, the negative reciprocal of that will be the slope of the normal, and we can then use point-slope form, $y-y_1=m(x-x_1),$ to find the equation.

Edit: If you want to check your work, hover over the box below for the solution:

$$\begin{align*}4x^2+xy^2-3y^3&=56 \\8x+\left(y^2+2xy\frac{dy}{dx}\right) - 9y^2\frac{dy}{dx}&=0 \\ 2xy\frac{dy}{dx} - 9y^2 \frac{dy}{dx} &= -8x-y^2 \\ \frac{dy}{dx}(2xy-9y^2) &= -8x-y^2 \\ \frac{dy}{dx} &= \frac{-8x-y^2}{2xy-9y^2} \end{align*} $$
Plug in $(-5,2)$:
$$ \begin{align*} \left.\frac{dy}{dx}\right]_{(-5,2)} &= \frac{-8(-5)-(2)^2}{2(-5)(2)-9(2)^2} \\ &= -\frac{9}{14} \end{align*} $$
Thus the slope of the tangent is $-\frac{9}{14}$, so the slope of the normal is $\frac{14}{9}$, and the equation of the normal is $y-2=\frac{14}{9}(x+5)$.

Answer 2

The derivative is $8x+y^2+2yx\dfrac{dy}{dx}-9y^2\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx}=\dfrac{-y^2-8x}{2yx-9y^2}$

Substituting the point $(-5,2)$ we have $\dfrac{dy}{dx}=(-4+40)/(-20-36)=-\frac{9}{14}$

Since we want the normal line, we take the inverse reciprocal of the slope: $\frac{14}{9}$.

We can use the point slope form to find the line: $y-2=\frac{14}{9}(x+5)$