<s>[INST] 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.
 [/INST] 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:
$$(a\cos\alpha)^n + (b\sin\alpha)^n = p^n$$
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