Question

The condition that the straight line $lx+my=n$ may be a normal to the hyperbola $b^2{x}^2-a^2{y}^2=a^2{b}^2$ is given by

• A. $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{(a^2+b^2{)}^2}{n^2}$
• B. $\frac{l^2}{a^2}-\frac{m^2}{b^2}=\frac{(a^2+b^2{)}^2}{n^2}$
• C. $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2-b^2{)}^2}{n^2}$
• D. $\frac{l^2}{b^2}+\frac{m^2}{b^2}=\frac{(a^2-b^2{)}^2}{n^2}$

Answer 1

The correct option is A $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{(a^2+b^2{)}^2}{n^2}$

Any normal to the hyperbola $b^2{x}^2-a^2{y}^2=a^2{b}^2$ is $\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^2+b^2\cdots \left(1\right)$
But the given normal equation is $lx+my=n\cdots \left(2\right)$

On comparing $\left(1\right)$ and $\left(2\right)$
$\Rightarrow \frac{l}{\frac{a}{\sec \theta }}=\frac{m}{\frac{b}{\tan \theta }}=\frac{n}{a^2+b^2}$
$\Rightarrow \sec \theta =\frac{a}{l}\left(\frac{n}{a^2+b^2}\right)$
and $\tan \theta =\frac{b}{m}\left(\frac{n}{a^2+b^2}\right)$

Using the identity ${\sec }^2\theta -{\tan }^2\theta =1$, we get
$\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{(a^2+b^2{)}^2}{n^2}$