JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Ellipse

For the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$, the tangent with slope $m$ is:

$y=mx\pm \sqrt{16m^2+9}$

For the circle $x^2+y^2=12$, the tangent with slope $m$ is:

$y=mx\pm \sqrt{12(1+m^2)}$

For a common tangent, the constant terms must be equal:

$\sqrt{16m^2+9}=\sqrt{12(1+m^2)}$

Squaring both sides:

$16m^2+9=12+12m^2$

$4m^2=3$

$m^2=\frac{3}{4}$

Therefore, $12m^2=12\cdot \frac{3}{4}=9$.