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. $y=2\sqrt{\frac{3}{7}}x+15\sqrt{17}$
• D. None of these

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 equation of common tangent be
$y=mk+c$

Since, $y=mx+c$ is a tangent to $\frac{x^2}{16}-\frac{y^2}{9}=1$

$\therefore c^2=16m^2-9\dots \left(i\right)[c^2=a^2{m}^2-b^2]$

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

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

From Eqs(i) and (ii) we get

$16m^2-9=9+9m^2\Rightarrow m^2=\frac{18}{7}\Rightarrow m=3\sqrt{\frac{2}{7}}$

From Eq.(ii) we get

$c^2=9(1+\frac{18}{7})\Rightarrow c=\frac{\pm 15}{\sqrt{7}}$

Hence $y=3\sqrt{\frac{2}{7}}x+\frac{15}{\sqrt{7}}$