JEE Main 2022 (Online) 26th June Evening ShiftQuestion 1
If $m$ is the slope of a common tangent to the curves $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ and $x^{2}+y^{2}=12$, then $12m^{2}$ is equal to :
Ellipse
Solution
Tangent $y=mx+c$ to ellipse: $c^2=16m^2+9$; to circle: $c^2=12(1+m^2)$. $16m^2+9=12+12m^2\Rightarrow m^2=\tfrac34$. $12m^2=9$. Option (B).