Let L be a common tangent


Let $L$ be a common tangent line to the curves $4 x^{2}+9 y^{2}=36$ and $(2 x)^{2}+(2 y)^{2}=31$. Then the square of the slope of the line $L$ is


Given curves are $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$


let slope of common tangent be $\mathrm{m}$

so tangents are $\mathrm{y}=\mathrm{mx} \pm \sqrt{9 \mathrm{~m}^{2}+4}$

$\mathrm{y}=\mathrm{mx} \pm \frac{\sqrt{31}}{2} \sqrt{1+\mathrm{m}^{2}}$

hence $9 \mathrm{~m}^{2}+4=\frac{31}{4}\left(1+\mathrm{m}^{2}\right)$

$\Rightarrow 36 \mathrm{~m}^{2}+16=31+31 \mathrm{~m}^{2} \Rightarrow \mathrm{m}^{2}=3$

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