Question

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 $k=$

• A. 144
• B. -9
• C. 25
• D. -21

Answer 1

Answer: (D)

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, $\frac{x^2}{4}+\frac{y^2}{9}=1$
$\Rightarrow \frac{2x}{4}+\frac{2y{y}^{\prime }}{9}=0$
$\Rightarrow y^{\prime }=-\frac{9}{4}\frac{x}{y}$ ...(ii)
Again, $\frac{x^2}{16}-\frac{y^2}{k}=1$
$\Rightarrow \frac{2x}{16}-\frac{2y{y}^{\prime }}{k}=0$
$\Rightarrow y^{\prime }=\frac{k}{16}\frac{x}{y}$ ...(iii)
Since both curves are orthogonal
$\therefore \frac{-9}{4}\frac{x}{y}\times \frac{k}{16}\frac{x}{y}=-1$
$\Rightarrow 9k{x}^2=64y^2$
From Eq. (i), we have
$9k\left(\frac{144+16k}{36+k}\right)=64\left(\frac{-27k}{36+k}\right)\Rightarrow k=-21$