Question on tangent and normal for the curve ${x^3 \over a}+{y^3 \over b} =xy$

Question

Find at what point on the curve $${x^3 \over a}+{y^3 \over b} =xy,$$ the tangent is parallel to one of the coordinate axes.

Answer 1

It would be improper to provide you with a full solution as you haven't included details about your attempt at the solution. I hope this partial solution helps you as a guide and you can fill out the details on your own.

Take a derivative with respect to x on both sides of the equation:

$\frac{d}{dx} (\frac{x^3}{a}+ \frac{y^3}{b}) = \frac{d}{dx}(xy)$

You can make use of chain rule here (https://en.wikipedia.org/wiki/Chain_rule) which gives rise to the $\frac{dy}{dx}$ term in the equation. Based on your comments, I do not think you'll have any trouble with this part. I'll omit the denominator part for now.

We get : $\frac{dy}{dx} = \frac{\frac{3x^2}{a}-y}{...}$

For the tangent to be parallel to the x-axis, $\frac{dy}{dx}$ must be $0$. This gives:

$\frac{3x^2}{a}-y = 0 \Longrightarrow y=\frac{3x^2}{a}$

You can now plug this into the original equation of the curve to get $x=\sqrt[3]{\frac{2a^2b}{27}}$. You can also get the y-value from this.

Can you compute the remaining part by yourself now?