Question

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

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

Answer 1

Answer: (B), (D)

$\sqrt{7}y=-3\sqrt{2}x-15$

Let $y=mx+c$ be a common tangent to $9x^2-16y^2=144$ and
$x^2+y^2=9$

As $y=mx+c$ is a tangent to $\frac{x^2}{16}-\frac{y^2}{9}=1$, so
$c^2=a^2{m}^2-b^2\Rightarrow c^2=16m^2-9\cdots \left(1\right)$

As $y=mx+c$ is a tangent to $x^2+y^2=9,$ so
$\left|\frac{c}{\sqrt{1+m^2}}\right|=3\Rightarrow c^2=9(1+m^2)\cdots \left(2\right)$
From equations $\left(1\right)$ and $\left(2\right)$, we get
$16m^2-9=9+9m^2\Rightarrow m^2=\frac{18}{7}\Rightarrow m=\pm \frac{3\sqrt{2}}{\sqrt{7}}\Rightarrow c^2=\frac{225}{7}\Rightarrow c=\pm \frac{15}{\sqrt{7}}$

Therefore, the equation of the common tangents are
$y=\pm \frac{3\sqrt{2}}{\sqrt{7}}x\pm \frac{15}{\sqrt{7}}\Rightarrow \sqrt{7}y=\pm 3\sqrt{2}x\pm 15$
Hence, the common tangent equations are
$\sqrt{7}y=3\sqrt{2}x+15\sqrt{7}y=-3\sqrt{2}x-15$