Discussion Forum

 If the curves $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $\frac{x^{2}}{16}-\frac{y^{2}}{k}=1$ cut each other orthogonally , then k=

Answer:

Explanation:

 We have,

$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $\frac{x^{2}}{16}-\frac{y^{2}}{k}=1$ 

 On solving  these equation , we get

 $x^{2}=\frac{144+16k}{36+k}$ and $y^{2}= \frac{-27k}{36+k}$  ..........(i)

 Now,     $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$

 $\Rightarrow$    $\frac{2x}{4}+\frac{2yy'}{9}=0\Rightarrow y'=-\frac{9}{4}\frac{x}{y}$    ..........(ii)

 again ,       $\frac{x^{2}}{16}-\frac{y^{2}}{k}=1 \Rightarrow \frac{2x}{16}-\frac{2yy'}{k}=0$

$\Rightarrow$        $y'= \frac{k}{16} \frac{x}{y}$          ............(iii)

 Since both curves are orthogonal

 $\therefore$    $\frac{-9}{4}\frac{x}{y}\times\frac{k}{16}\frac{x}{y}=-1\Rightarrow 9kx^{2}=64y^{2}$

 From Eq.(i) , we have

$9k\left(\frac{144+16k}{36+k}\right)=64\left(\frac{-27k}{36+k}\right)\Rightarrow k=-21$