Question

If the line $x\cos \alpha +y\sin \alpha =p$ be tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then?

Answer 1

$x\cos \alpha +y\sin \alpha =p$

$\Rightarrow y=\frac{1^2-x\cos \alpha }{\sin \alpha }⟶\left(i\right)$

$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$b^2{x}^2+a^2{y}^2=a^2{b}^2$

$b^2{x}^2+a^2{(p\csc \alpha -x\cot \alpha )}^2=a^2{b}^2$

$\Rightarrow x^2[b^2+a^2{\cot }^2\alpha ]-2a^2\left(p\csc \alpha \cot \alpha \right)x+a^2(p^2{\csc }^{2}x-6^2)=0\to \left(ii\right)$

If $x\cos \alpha +y\sin \alpha =p$ is tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ then equation $\left(ii\right)$ has equal roots.

$\therefore 4a^4{p}^2{\csc }^2\alpha -4(b^2+0^2{\cot }^2\alpha )(a^2{p}^2{\csc }^2\alpha -a^2{b}^2)=0$

$\Rightarrow b^2{p}^2{\csc }^2\alpha -b^4-a^2{b}^2{\cot }^2\alpha =0$

$\Rightarrow p^2-b^2{\sin }^2\alpha -a^2{\cos }^2\alpha =0$

$\Rightarrow p^2=a^2{\cos }^2\alpha +b^2{\sin }^2\alpha$

Hence, solved.