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