I have this question: If the line $x\cos\alpha + y\sin\alpha = p$ touches the curve $\left(\frac{x}{a}\right)^\frac{n}{n - 1} + \left(\frac{y}{b}\right)^\frac{n}{n - 1} = 1$
then prove that $\left(a\cos\alpha\right)^n + \left(b\sin\alpha\right)^n = p^n$
I know that the equation given is an equation of the line in normal form with perpendicular distance $p$ from origin.
Also the slope of given line is $-\cot\alpha$
and this slope of line will be equal to the slope of the curve. But equating both is not yielding the desired result.
The only little progress I seem to make after substituting $x$ and $y$ from the equation of line to the equation of curve seems futile to prove this further.
I seem to make no further progress in this question. What should I do?
Note the slope of the line is $$ m_1=-\frac{\cos\alpha}{\sin\alpha} $$ and the slope of the tangent line of the curve at $(x_0,y_0)$ is $$ m_2=y'|_{x=x_0}=-\frac{b}{a}\left(\frac{bx_0}{ay_0}\right)^{\frac1{n-1}}.$$ Thus from $m_1=m_2$ we have $$ y_0=\frac{b^n\sin^{n-1}\alpha}{a^n\cos^{n-1}\alpha}x_0.$$ Noting that $(x_0,y_0)$ is on the line, we have $$ x_0=\frac{pa^n\cos^{n-1}\alpha}{a^n\cos^{n}\alpha+b^n\sin^{n}\alpha}, y_0=\frac{pb^n\sin^{n-1}\alpha}{a^n\cos^{n}\alpha+b^n\sin^{n}\alpha}. $$ But $(x_0,y_0)$ is on the curve, namely $\left(\frac{x_0}{a}\right)^\frac{n}{n - 1} + \left(\frac{y_0}{b}\right)^\frac{n}{n - 1} = 1$ from which we deduce $$\left(a\cos\alpha\right)^n + \left(b\sin\alpha\right)^n = p^n. $$