Question

If the straight line $x\cos \alpha +y\sin \alpha =p$ touches the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ then prove that $a^2{\cos }^2\alpha +b^2{\sin }^2\alpha =p^2$

Answer 1

$x\cos \alpha +y\sin \alpha =P$ touches $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

then also at that point should be same

$\to \cos \alpha +m\sin \alpha =0$ or $m=\cot \alpha ---\left(1\right)$

$\to \frac{2x}{a^2}+\frac{2ym}{b^2}=0$ or $x=a^2\frac{\cot \alpha y}{b^2}---\left(2\right)$

put $\left(2\right)$ in $\left(1\right)$

$y(\frac{a^2{\cos }^2\alpha }{b^2\sin \alpha }+\sin \alpha )=P$

$\Rightarrow \frac{y}{b^2\sin \alpha }[a^2{\cos }^2\alpha +b^2{\sin }^2\alpha ]=P$

or $y=\frac{P{b}^2\sin \alpha }{a^2{\cos }^2\alpha +b^2{\sin }^2\alpha }\Rightarrow x=P{a}^2\cos \alpha \left(\frac{1}{a^2{\cos }^2\alpha +b^2{\sin }^2\alpha }\right)----\left(3\right)$

Put $\left(3\right)$ back in $\left(2\right)$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow \frac{P^2{a}^4{\cos }^2\alpha }{(a^2{\cos }^2\alpha +b^2{\sin }^2\alpha {)}^2{a}^2}+\frac{P^4{b}^2{\sin }^2\alpha }{b^2(a^2{\cos }^2\alpha +b^2{\sin }^2\alpha {)}^2}=1$

$\Rightarrow P^2(a^2{\cos }^2\alpha +b^2{\sin }^2\alpha )=(a^2{\cos }^2\alpha +b^2{\sin }^2\alpha {)}^2$

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