Question

If the line $lx+my-n=0$ will be a normal to the hyperbola, then $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{(a^2+b^2{)}^2}{k}$, where k is equal to

• A. $n$
• B. $n^2$
• C. $n^3$
• D. None of these

Answer 1

The correct option is B $n^2$

The equation of any normal to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $axcos\varphi +bycot\varphi =a^2+b^2$
$\Rightarrow axcos\varphi +bycot\varphi -(a^2+b^2)=0$ ...... (i)
The straight line $lx+my-n=0$ will be normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Then eq. (i) and $lx+my-n=0$ represents the same line,

$\therefore l=acos\varphi ,m=bcos\varphi ,n=(a^2+b^2)$

$\therefore \frac{acos\varphi }{l}=\frac{bcot\varphi }{m}=\frac{a^2+b^2}{n}$

$\Rightarrow sec\varphi =\frac{na}{l(a^2+b^2)}$
and $tan\varphi =\frac{nb}{m(a^2+b^2)}$

$\because se{c}^2\varphi -ta{n}^2\varphi =1$
$⟹\frac{n^2{a}^2}{l^2(a^2+b^2{)}^2}-\frac{n^2{b}^2}{m^2(a^2+b^2{)}^2}=1$

$\Rightarrow \frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{(a^2+b^2{)}^2}{n^2}$
But given equation of normal is
$\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{(a^2+b^2{)}^2}{k}$

$\therefore k=n^2$.