Discussion Forum

If the  line x+y+k=0 is a normal to the hyperbola  $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$, then k

Answer:

Explanation:

 We know that  equation of normal of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is 

   $\frac{a^{2}x}{x_{1}}+\frac{b^{2}y}{y_{1}}=a^{2}-b^{2}$

$\therefore$      Equations of normal to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$

 is 

  $\frac{9x}{x_{1}}+\frac{4y}{y_{1}}=9+4$

 $\Rightarrow$   $\frac{9x}{x_{1}}+\frac{4y}{y_{1}}=13$

  Since,  line x+y=k is normal to the given hyperbola 

 $\therefore$    $\frac{\frac{9}{x_{1}}}{1}=\frac{\frac{4}{y_{1}}}{1}=\frac{13}{k}$

 $\Rightarrow$     $\frac{9}{x_{1}}=\frac{4}{y_{1}}$

   = $\frac{13}{k}$

    $\Rightarrow$      $x_{1}= \frac{9k}{13}$

                $y_{1}=\frac{4k}{13}$

 Since ($(x_{1},y_{1})$  lie on the hyperbola

$\therefore$    $\frac{\left(\frac{9k}{13}\right)^{2}}{9}-\frac{\left(\frac{4k}{13}\right)^{2}}{9}=1$

  $\Rightarrow$       $\frac{9k^{2}}{169}-\frac{4k^{2}}{169}=1$

 $\Rightarrow$    $5k^{2}=169$

   $\Rightarrow$   $k=\pm\frac{13}{\sqrt{5}}$