Question

The equation of common tangent to the hyperbola $9x^2-16y^2=144$ and circle $x^2+y^2=9$ is

• A. $y=\sqrt{\frac{6}{7}}x+\frac{15}{\sqrt{7}}$
• B. $y=3\sqrt{\frac{2}{7}}x+\frac{15}{\sqrt{7}}$
• C. $y=3\sqrt{\frac{2}{7}}x+15\sqrt{7}$
• D. $y=\sqrt{\frac{6}{7}}x+15\sqrt{7}$

Answer 1

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

Let $y=mx+c$ be a common tangent to $9x^2-16y^2=144$ and
$x^2+y^2=9$
Since, $y=mx+c$ is a tangent to $\frac{x^2}{16}-\frac{y^2}{9}=1,$ therefore
$c^2=a^2{m}^2-b^2$
$\Rightarrow c^2=16m^2-9\dots \left(1\right)$
Since, $y=mx+c$ is a tangent to $x^2+y^2=9,$ therefore
$\left|\frac{c}{\sqrt{1+m^2}}\right|=3$
$\Rightarrow c^2=9(1+m^2)\dots \left(2\right)$
From equations $\left(1\right)$ and $\left(2\right)$, we get
$16m^2-9=9+9m^2$
$\Rightarrow m=\pm 3\sqrt{\frac{2}{7}}$
By equation $\left(1\right)$, we get $c=\pm \frac{15}{\sqrt{7}}$

Hence, $y=3\sqrt{\frac{2}{7}}x+\frac{15}{\sqrt{7}}$ is the equation of common tangent.