Q68 / 120 · VITEEE 2014 · Mathematics
If the line $l x+m y-n=0$ will be a normal to the hyperbola, then $\frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{k}$, where $k$ is equal to
- A$n$
- Correct answerB$n^{3}$
- C$n^{2}$
- DNone of these
Solution
The equation of any normal
$\begin{aligned}
& \text { to } \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text { is } a x \cos \phi+b y \cot \phi \\
&=a^{2}+b^{2} \\
\Rightarrow & a x \cos \phi+b y \cot \phi-\left(a^{2}+b^{2}\right)=0 \ldots \ldots \text { (i) }
\end{aligned}$
The straight line $l x+m y-n=0$ will be normal to the hyperbola
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, then eq. (i) and $l x+m y-n=$ 0 represent the same line,
$\begin{array}{l}
\therefore \frac{a \cos \phi}{l}=\frac{b \cot \phi}{m}=\frac{a^{2}+b^{2}}{n} \\
\Rightarrow \sec \phi=\frac{n a}{l\left(a^{2}+b^{2}\right)} \\
\text { and } \tan \phi=\frac{n b}{m\left(a^{2}+b^{2}\right)} \\
\frac{n^{2} a^{2}}{l^{2}\left(a^{2}+b^{2}\right)^{2}}-\frac{n^{2} b^{2}}{m^{2}\left(a^{2}+b^{2}\right)^{2}}=1 \\
{\left[\because \sec ^{2} \phi-\tan ^{2} \phi=1\right]}
\end{array}$
$\Rightarrow \frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{n^{2}}$
But given equation of normal is
$\begin{array}{l}
\frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{k} \\
\therefore k=n^{2}
\end{array}$
Hence, the correct answer is option: 'B'.