Question

The line $lx+my+n=0$ will be a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ if, :

• A. $\frac{a^2}{l^2}+\frac{b^2}{m^2}=1$
• B. $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{{(a^2+b^2)}^2}{n^2}$
• C. $am+bl=0$
• D. $\frac{a^2}{l^2}-\frac{b^2}{m^2}=0$

Answer 1

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

The equation of the normal at $\left(a\sec \varphi ,b\tan \varphi \right)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
$ax\sin \varphi +by=(a^2+b^2)\tan \varphi \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(i\right)$
And the equation of the line is
$lx+my+n=0\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(ii\right)$
Now $\left(i\right)$ and $\left(ii\right)$ represent the same line
$\therefore \frac{a\sin \varphi }{l}=\frac{b}{m}=\frac{(a^2+b^2)\tan \varphi }{-na}$
$=\frac{(a^2+b^2)\sin \varphi }{-n\cos \varphi }$
$\Rightarrow \sin \varphi =\frac{bl}{am}\cos \varphi =\frac{(a^2+b^2)l}{-na}$
$1=\frac{b^2{l}^2}{a^2{m}^2}+\frac{{(a^2+b^2)}^2}{n^2{a}^2}\left[squaringandadding\right]$
$\Rightarrow \frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{{(a^2+b^2)}^2}{n^2}$
Hence, option 'B' is correct.