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

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(A) The given curves are $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ (an ellipse) and $x^{2} + y^{2} = \frac{31}{4}$ (a circle).Let the slope of the commo

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$3$

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(A) The given curves are $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ (an ellipse) and $x^{2} + y^{2} = \frac{31}{4}$ (a circle).Let the slope of the common tangent be $m$.The equation of a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $y = mx \pm \sqrt{a^{2}m^{2} + b^{2}}$.Here $a^{2} = 9$ and $b^{2} = 4$,so the tangent is $y = mx \pm \sqrt{9m^{2} + 4}$.The equation of a tangent to the circle $x^{2} + y^{2} = r^{2}$ is $y = mx \pm r\sqrt{1 + m^{2}}$.Here $r^{2} = \frac{31}{4}$,so the tangent is $y = mx \pm \frac{\sqrt{31}}{2}\sqrt{1 + m^{2}}$.For the lines to be identical,the constant terms must be equal:$9m^{2} + 4 = \frac{31}{4}(1 + m^{2})$.Multiplying by $4$,we get $36m^{2} + 16 = 31 + 31m^{2}$.$5m^{2} = 15$.$m^{2} = 3$.

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

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