Answer
$x^{\frac 23}+y^{\frac 23}=a^{\frac 23}$
$\frac 23x^{-\frac 13}+\frac 23y^{-\frac 13}y'=0$
$y'=-(\frac yx)^{\frac 13}$
Let $x=h,y=k$
So at point $(h,k)$,$y'=-(\frac kh)^{\frac 13}$
So the tangent at $(h,k)$ is $y-k=-(\frac kh)^{\frac 13}(x-h)$
At y-axis,$x=0,y=k+k^{\frac 13}h^{\frac 23}$
At x-axis,$y=0,x=h+h^{\frac 13}k^{\frac 23}$
So the length=$(k+k^{\frac 13}h^{\frac 23}-0)^2+(0-(h+h^{\frac 13}k^{\frac 23}))^2=a^{\frac 43}(k^{\frac 23}+h^{\frac 23})$
Because $x^{\frac 23}+y^{\frac 23}=a^{\frac 23}$
So the length=$a^{\frac 23}a^{\frac 43}=a^2$
Work Step by Step
$x^{\frac 23}+y^{\frac 23}=a^{\frac 23}$
$\frac 23x^{-\frac 13}+\frac 23y^{-\frac 13}y'=0$
$y'=-(\frac yx)^{\frac 13}$
Let $x=h,y=k$
So at point $(h,k)$,$y'=-(\frac kh)^{\frac 13}$
So the tangent at $(h,k)$ is $y-k=-(\frac kh)^{\frac 13}(x-h)$
At y-axis,$x=0,y=k+k^{\frac 13}h^{\frac 23}$
At x-axis,$y=0,x=h+h^{\frac 13}k^{\frac 23}$
So the length=$(k+k^{\frac 13}h^{\frac 23}-0)^2+(0-(h+h^{\frac 13}k^{\frac 23}))^2=a^{\frac 43}(k^{\frac 23}+h^{\frac 23})$
Because $x^{\frac 23}+y^{\frac 23}=a^{\frac 23}$
So the length=$a^{\frac 23}a^{\frac 43}=a^2$
