Discussion Forum

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

Answer:

Explanation:

 The equation of any normal

to $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is  $a x \cos \phi+by \cot \phi=a^{2}+b^{2}$

 $\Rightarrow$   $a x \cos \phi+by \cot \phi -(a^{2}+b^{2})=0....(i)$

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 represent  the same line, 

 $\therefore$   $\frac{a \cos \phi}{l}=\frac{b \cot \phi}{m}=\frac{a^{2}+b^{2}}{n}$

$\Rightarrow$  $\sec \phi= \frac{na}{l(a^{2}+b^{2})}$

  and $\tan \phi = \frac{nb}{m(a^{2}+b^{2})}$ 

$\frac{n^{2}a^{2}}{l^{2}(a^{2}+b^{2})^{2}}-\frac{n^{2}b^{2}}{m^{2}(a^{2}+b^{2})^{2}}=1$

                                                   $[ \because   \sec^{2} \phi-\tan^{2} \phi=1]$

$\frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{n^{2}}$

But given equation  of normal is 

$\frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{k}$

$\therefore$   $k=n^{2}$