Let L be a common tangent line to the curves | vedclass

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Question

Given curves are $\frac{ x ^{2}}{9}+\frac{ y ^{2}}{4}=1$

$x^{2}+y^{2}=\frac{31}{4}$

let slope of common tangent be $m$

so tangents are $y=m x

Vedclass · Accepted Answer

$3$

Vedclass · Answer

Given curves are $\frac{ x ^{2}}{9}+\frac{ y ^{2}}{4}=1$

$x^{2}+y^{2}=\frac{31}{4}$

let slope of common tangent be $m$

so tangents are $y=m x \pm \sqrt{9 m^{2}+4}$

$y=m x \pm \frac{\sqrt{31}}{2} \sqrt{1+m^{2}}$

hence $9 m ^{2}+4=\frac{31}{4}\left(1+ m ^{2}\right)$

$\Rightarrow 36 m ^{2}+16=31+31 m ^{2} \Rightarrow m ^{2}=3$

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 ..... .

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