Question

Find the common tangent to the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and an ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$.

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

Answer 1

The correct option is D $y=\pm x\pm \sqrt{7}$

Equation of tangent to the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ is,

$y=mx\pm \sqrt{a^2{m}^2-b^2}\Rightarrow y=mx\pm \sqrt{16m^2-9}....\left(1\right)$

And equation of tangent to the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ is,

$y=mx\pm \sqrt{a^2{m}^2+b^2}\Rightarrow y=mx\pm \sqrt{4m^2+3}....\left(2\right)$

Since both lines are same

$\Rightarrow 16m^2-9=4m^2+3\Rightarrow m^2=1\Rightarrow m=\pm 1$

Hence required tangent is, $y=\pm x\pm \sqrt{7}$