How to prove that $(a\cos\alpha)^n + (b\sin\alpha)^n = p^n$ under the following conditions? How to prove that $(a\cos\alpha)^n + (b\sin\alpha)^n = p^n$ when then line $x\cos\alpha + y\sin\alpha = p$ touches the curve  $$\left (\frac{x}{a} \right )^\frac{n}{n-1} + \left (\frac{y}{b} \right )^\frac{n}{n-1}=1$$
What I've tried:
I've equated the derivative of the given line with the general derivative of the given curve but couldn't proceed to any meaningful step thereafter.

At the tangent point (x, y), the normal of the two curves are parallel, so we can get
 $$\{\cos \alpha, \sin \alpha \}//\{\frac{x^\frac{1}{n-1}}{a^\frac{n}{n-1}}, \frac{x^\frac{1}{n-1}}{a^\frac{n}{n-1}}\} $$
so we get the equation:
$$\frac{x^\frac{1}{n-1}}{a^\frac{n}{n-1}\cos \alpha}=\frac{x^\frac{1}{n-1}}{a^\frac{n}{n-1}\sin \alpha}=k$$
Hence:
$$x=k^{n-1}a^n(\cos\alpha)^{n-1}$$ $$y=k^{n-1}b^n(\sin\alpha)^{n-1}$$
Take x and y into $x\cos\alpha + y\sin\alpha = p$ we get:
$$k^{n-1}((a\cos\alpha)^n+(b\sin\alpha)^n)=p$$
Take x and y into $\left (\frac{x}{a} \right )^\frac{n}{n-1} + \left (\frac{y}{b} \right )^\frac{n}{n-1}=1$ we get:
$$k^n((a\cos\alpha)^n+(b\sin\alpha)^n)=1$$
Therefore:$$k=1/p$$
Take K into $k^n(a^n\cos\alpha^n+b^n\sin\alpha^n)=1$  get the result:
$$(a\cos\alpha)^n + (b\sin\alpha)^n = p^n$$
