Question

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

Answer 1

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

tangent:$x\cos \alpha +y\sin \alpha =p---\left(i\right)$

$\Rightarrow y=\frac{-\left(\cos \alpha \right)}{\sin \alpha }x+\frac{p}{\sin \alpha }$

for, $\left(i\right)$ being tangent,

$c=\pm \sqrt{a^2{m}^2+b^2}$

$\Rightarrow (\frac{p}{\sin \alpha }{)}^2=(a^2(\frac{-\left(\cos \alpha \right)}{\sin \alpha }{)}^2+b^2)$

$\Rightarrow \frac{p^2}{{\sin }^2\alpha }=\frac{a^2\left(\cos {}^2\alpha \right)}{\sin {}^2\alpha }+b^2$

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