Question

Common tangent equations to $9x^2-16y^2=144$ and $x^2+y^2=9$ are

• A. $y=\pm \frac{3x}{\sqrt{7}}+\pm \frac{15}{\sqrt{7}}$
• B. $y=\pm 3\sqrt{\frac{2}{7}}x\pm \frac{15}{\sqrt{7}}$
• C. $y=\pm 2\sqrt{\frac{2}{7}}x\pm \frac{15}{\sqrt{7}}$
• D. $y=\pm \frac{4x}{\sqrt{7}}\pm \frac{15}{\sqrt{7}}$

Answer 1

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

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 $9x^2-16y^2=144,$ 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 common tangent equations are
$y=\pm 3\sqrt{\frac{2}{7}}x\pm \frac{15}{\sqrt{7}}$