Question

If the line $xcos\alpha +ysin\alpha =p$ be normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , then

• A. $p^2\left(a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha \right)=a^2-b^2$
• B. $p^2\left(a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha \right)={\left(a^2-b^2\right)}^2$
• C. $p^2\left(a^{2}se{c}^2\alpha +b^{2}cose{c}^2\alpha \right)=a^2-b^2$
• D. $p^2\left(a^{2}se{c}^2\alpha +b^{2}cose{c}^2\alpha \right)={\left(a^2-b^2\right)}^2$

Answer 1

Answer: (D)

The equation of any normal to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$ax\sec \varphi -by\mathrm{cosec}\varphi =a^2-b^2...(i)$
The straight line $x\cos \alpha +y\sin \alpha =p$ will be a normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ if Eq. (i) and $x\cos \alpha +y\sin \alpha =p$ represent the same line.
$\therefore \frac{a\sec \varphi }{\cos \alpha }=\frac{-bcosec\varphi }{\sin \alpha }=\frac{a^2-b^2}{p}$
$\Rightarrow \cos \varphi =\frac{ap}{\left(a^2-b^2\right)\cos \alpha }$
$\sin \varphi =\frac{-bp}{\left(a^2-b^2\right)\sin \alpha }$
$\because {\sin }^2\varphi +{\cos }^2\varphi =1$
$\Rightarrow \frac{b^2{p}^2}{{\left(a^2-b^2\right)}^2{\sin }^2\alpha }+\frac{a^2{p}^2}{{\left(a^2-b^2\right)}^2{\cos }^2\alpha }=1$
$\Rightarrow p^2\left(b^2{cosec}^2\alpha +a^2{\sec }^2\alpha \right)={\left(a^2-b^2\right)}^2$