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. $y=2\sqrt{\frac{2}{7}}+15\sqrt{7}$
• D. None of the above

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 $9x^2-16y^2=144$

$\therefore$ $c^2=a^2{m}^2-b^2$ $\Rightarrow$

$c^2=16m^2-\mathrm{9.......}\left(i\right)$

Now, $y=mx+c$ is a tangent to $x^2+y^2=9$

$\therefore$ $\frac{c}{\sqrt{m^2+1}}=3$ $\Rightarrow c=9(1+m^2)......\left(ii\right)$

From Eqs. $\left(i\right)$ and $\left(ii\right)$, we get

$16m^2-9=9+9m^2$

$\Rightarrow$ $m=\pm 3\sqrt{\frac{2}{7}}$

On putting the value of $m$ in Eq $\left(ii\right)$ we get

$c=\pm \frac{15}{\sqrt{7}}$

Hence $y=3\sqrt{\frac{2}{7}}x+\frac{15}{\sqrt{7}}$ is a common tangent