Find the condition on $a$ and $b$ such that the line $\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve $x^{2/3}+y^{2/3}=1$
I'm a bit weak at coordinate geometry so what I tried to do was differentiate to get the derivitive of y with respect to x and some how equate that to the slope of the given line. But this didn't work out.
Firstly you can see a good thing happening in this question. That is, we can substitute as follows:
$x^\frac{1}{3}$=$\sin{\theta}$
$y^\frac{1}{3}$=$\cos{\theta}$
$x$=$\sin^3{\theta}$
$y$=$\cos^3{\theta}$
$\frac{dx}{d\theta}$ =$3\sin^2{\theta}$$\cos{\theta}$
$\frac{dy}{d\theta}$ =$-3\sin{\theta}$$\cos^2{\theta}$
$\frac{dy}{dx}$=$-\cot{\theta}$
Therefore,
$y-cos^3{\gamma}$=$-\cot{\gamma}${$x-\sin^3{\gamma}$}
Is the tangent equation.
{here ${\gamma}$=random taken value of ${\theta}$ which is the parameter at the tangency point}
Now,
Put $(a,0)$ and $(0,b)$
You will get
$a$=$\sin{\gamma}$ $[\sin^2{\gamma}$+$\cos^2{\gamma}$]
=$\sin{\gamma}$
Similarly, you will get $b=\cos{\gamma}$
Therefore,
$a^2+b^2=1$
is the required condition.