If a tangent of slope 2 to the ellipse frac{x | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. For a line $y=mx+c$ to be a tangent to the ellipse,$c^2=a^2m^2+b^2$. Given slo

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$5$

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(C) The equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. For a line $y=mx+c$ to be a tangent to the ellipse,$c^2=a^2m^2+b^2$. Given slope $m=2$,the tangent equation is $y=2x \pm \sqrt{4a^2+b^2}$,which can be rewritten as $2x-y \pm \sqrt{4a^2+b^2}=0$. Since this line is tangent to the circle $x^2+y^2=4$ (radius $r=2$,center $(0,0)$),the perpendicular distance from the origin to the line must equal the radius: $\frac{|\pm \sqrt{4a^2+b^2}|}{\sqrt{2^2+(-1)^2}} = 2$. $\frac{\sqrt{4a^2+b^2}}{\sqrt{5}} = 2$ $\Rightarrow \sqrt{4a^2+b^2} = 2\sqrt{5}$ $\Rightarrow 4a^2+b^2 = 20$. Using the $AM \geq GM$ inequality: $\frac{4a^2+b^2}{2} \geq \sqrt{4a^2b^2} = 2ab$. $\frac{20}{2} \geq 2ab$ $\Rightarrow 10 \geq 2ab$ $\Rightarrow ab \leq 5$. Thus,the maximum value of $ab$ is $5$.

If a tangent of slope $2$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ touches the circle $x^2+y^2=4$,then the maximum value of $ab$ is

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