Find the length of the tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ which is intercepted between the axes.

Question

Find the length of the tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ which is intercepted between the axes.

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}\implies \frac{dy}{dx}={(\frac{-y}{x})}^{\frac{1}{3}}$.

Slope at $(p,q)$ will be ${(\frac{-q}{p})}^{\frac{1}{3}}$.

So equation of tangent at $(p,q);$ $y-q={(\frac{-q}{p})}^{\frac{1}{3}}(x-p)$.

How to find the length of tangent intercepted in between the axes from here?

Answer 1

Now, if $x=0$ we obtain $y=q+\sqrt[3]{qp^2}$ and for $y=0$ we obtain $x=p+\sqrt[3]{pq^2}.$

Thus, $A\left(p+\sqrt[3]{pq^2},0\right)$ and $B\left(0,q+\sqrt[3]{qp^2}\right)$ they are intersection points of the tangent with $x$-axis and with $y$- axis respectively.

Id est, $$AB=\sqrt{\left(p+\sqrt[3]{pq^2}\right)^2+\left( q+\sqrt[3]{qp^2}\right)^2}=$$ $$=\sqrt{p^{\frac{2}{3}}\left(p^{\frac{2}{3}}+q^{\frac{2}{3}}\right)^2+q^{\frac{2}{3}}\left(q^{\frac{2}{3}}+p^{\frac{2}{3}}\right)^2}= p^{\frac{2}{3}}+q^{\frac{2}{3}}=a^{\frac{2}{3}}.$$

Answer 2

Any point on $$x^{2/n}+y^{2/n}=a^{2/n}\ \ \ \ (1)$$ can be chosen as $(a\cos^nt,a\sin^nt)$

Differentiating $(1)$ wrt $x,$ $$\dfrac{dy}{dx}=-\dfrac{x^{(2-n)/n}}{y^{(2-n)/n}}$$

$$\dfrac{dy}{dx}_{\text{ at }(a\cos^nt,a\sin^nt)}=-\dfrac{x^{(2-n)/n}}{y^{(2-n)/n}}=-\dfrac{\cos^{2-n}t}{\sin^{2-n}t}$$

So, the equation of the tangent at $(a\cos^nt,a\sin^nt)$ will be $$\dfrac{y-a\sin^nt}{x-a\cos^nt}=-\dfrac{\cos^2t\sin^nt}{\sin^2t\cos^nt}$$

$$\iff x\cos^2t\sin^nt+y\sin^2t\cos^nt=a\cos^nt\sin^nt$$

$$\iff\dfrac x{\cos^{n-2}t}+\dfrac y{\sin^{n-2}t}=1$$

Here $n=3$

Can you take it from here?

See also: intercept form