I recently encountered this problem in my book:
If h and k be the intercepts on the coordinate axes of tangent to the curve $\left(\frac{x}{a}\right)^{\frac{2}{3}}+\left(\frac{y}{b}\right)^{\frac{2}{3}}=1$ at any point on it then prove that, $$\frac{h^2}{a^2}+\frac{k^2}{b^2}=1.$$
Now, the problem is to be solved by derivative. So, I differentiated the equation of the curve against x and set $$\left[\frac{dy}{dx}\right]_{{x_0},{y_0}}=\frac{y-y_0}{x-x_0}$$
This is the equation of the tangent at point ($\require{mhchem}\ce{x0,y0}$). Then I put $\ce{x=0, y=k}$ and then $\ce{x=h, y=0}$ in the equation of tangent to express $\ce{h}$ and $\ce{k}$ in terms of $\ce{x0}$ and $\ce{y0}$. Then I tried to put these values in $\mathrm{{\frac{h^2}{k^2}}+\frac{y^2}{b^2}}$. But the expression became very large and complicated.
Can you suggest an easier method?
[Note: As I am a high school student, so if you can provide a method that does not involve the use of many theorems would be more helpful. In other words, a proof involving less assumptions is better. But any suggestion is welcome.]
HINT
Note that
$$\left(\frac{x}{a}\right)^{\frac{2}{3}}+\left(\frac{y}{b}\right)^{\frac{2}{3}}=1\implies \frac23\frac1a\left(\frac{x}{a}\right)^{-\frac{1}{3}}dx+\frac23\frac1b\left(\frac{y}{b}\right)^{-\frac{1}{3}}dy=0 \implies \frac{dy}{dx}=-\left(\frac{b^4x}{a^4y}\right)^{-\frac13}$$