Question

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

Answer 1

Let the slope of tangent be $m$

The equation of tangent with slope $m$ to ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ is $y=mx\pm \sqrt{4m^2+3}$

The equation of tangent with slope $m$ to ellipse $\frac{x^2}{16}-\frac{y^2}{9}=1$ is $y=mx\pm \sqrt{16m^2-9}$

As the tangents are common to both the curves

$4m^2+3=16m^2-9⟹12m^2=12⟹m^2=1⟹m=\pm 1$

So The common tangent is $y=\pm x\pm \sqrt{7}$