Question

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

• A. $p^2(a^2{\cos }^2\alpha +b^2{\sin }^2\alpha )=a^2-b^2$
• B. $p^2(a^2{\cos }^2\alpha +b^2{\sin }^2\alpha )={(a^2-b^2)}^2$
• C. $p^2(a^2{\sec }^2\alpha +b^2{\csc }^2\alpha )=a^2-b^2$
• D. $p^2(a^2{\sec }^2\alpha +b^2{\csc }^2\alpha )={(a^2-b^2)}^2$

Answer 1

Answer: (D)

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

The equation of any normal to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$ax\sec \varphi -by\csc \varphi =a^2-b^2$ ......(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 equation (i) and $x\cos \alpha +y\sin \alpha =p$ represent the same line.
$\therefore \frac{a\sec \varphi }{\cos \alpha }=\frac{-b\csc \varphi }{\sin \alpha }=\frac{a^2-b^2}{p}$
$\Rightarrow \cos \varphi =\frac{ap}{(a^2-b^2)\cos \alpha }$
$\sin \varphi =\frac{-bp}{(a^2-b^2)\sin \alpha }$
$\because {\sin }^2\varphi +{\cos }^2\varphi =1$
$\Rightarrow \frac{b^2{p}^2}{{(a^2-b^2)}^2{\sin }^2\alpha }+\frac{a^2{p}^2}{{(a^2-b^2)}^2{\cos }^2\alpha }=1$
$\Rightarrow p^2(b^2{\csc }^2\alpha +a^2{\sec }^2\alpha )={(a^2-b^2)}^2$