Question

A common tangent to $9x^2-16y^2=144$ and $x^2+y^2=9$, is

• A. $y=\frac{3}{\sqrt{7}}x+\frac{15}{\sqrt{7}}$
• B. $y=3\sqrt{\frac{2}{7}}x+\frac{15}{\sqrt{7}}$
• C. $y=2\sqrt{\frac{3}{7}}x+15\sqrt{7}$
• D. $y=\frac{3}{\sqrt{7}}x-\frac{15}{\sqrt{7}}$

Answer 1

The correct option is B $y=3\sqrt{\frac{2}{7}}x+\frac{15}{\sqrt{7}}$

Given curves are $9x^2-16y^2=144$ and $x^2+y^2=9$.
Let the equation of common tangent be $y=mx+c$.
Since, $y=mx+c$ is a tangent to $x^2+y^2=9$,
$\left|\frac{0-0+c}{\sqrt{1+m^2}}\right|=3\Rightarrow c^2=9(1+m^2)\dots \left(1\right)$
Also, $y=mx+c$ is a tangent to $\frac{x^2}{16}-\frac{y^2}{9}=1$
$\Rightarrow c^2=16m^2-9\dots \left(2\right)$
(using condition of tangency)

From $\left(1\right)$ and $\left(2\right)$, we get
$16m^2-9=9+9m^2\Rightarrow m^2=\frac{18}{7}\Rightarrow m=\pm 3\sqrt{\frac{2}{7}}$
Also, $c=\pm \frac{15}{\sqrt{7}}$
Hence, $y=3\sqrt{\frac{2}{7}}x+\frac{15}{\sqrt{7}}$ is one of their common tangent.