Ellipse

Given, curves are $A: 4 x^{2}+9 y^{2}=36$ $B:(2 x)^{2}+(2 y)^{2}=31$ $L$ is common tangent line to curve $A$ and $B$. Let slope of $L$ be $m$. Given, curve $A$ is equation of ellipse. $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ Then, equation of tangent to ellipse will be $y=m x \pm \sqrt{9 m^{2}+4}$. . . (i) Curve $B$ is circle i.e $x^{2}+y^{2}=\frac{31}{4}$ $\text{ Then, equation of tangent to circle will be }$ $y=m x \pm \sqrt{ \left( \frac{31}{4} \right) m^{2}+ \left( \frac{31}{4} \right) }$ . . . (ii) On comparing Eqs. (i) and (ii), we get On comparing Eqs. (i) and (ii), we get $9 m^{2}+4= \left( \frac{31}{4} \right) m^{2}+ \left( \frac{31}{4} \right)$ $\Rightarrow 36 m^{2}+16=31 m^{2}+31 \Rightarrow 5 m^{2}=15$ $\Rightarrow m^{2}=3$ $\therefore$ Square of slope of line $L$ is 3 .