Question

An equilateral triangle is inscribed in an ellipse whose equation is $x^2+4y^2=4.$ One vertex of the triangle is $(0,1)$, one altitude is contained in the $y-$axis, and the length of each side is $\sqrt{\frac{m}{n}},$ where $m$ and $n$ are relatively prime positive integers. Then the value of $(m+n)$ is

Answer 1

$\frac{x^2}{4}+y^2=1\dots \left(1\right)$
Equation of side $AB$
$y-1=\sqrt{3}(x-0)$
$\Rightarrow y=\sqrt{3}x+1\dots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right),$
$\frac{x^2}{4}+(\sqrt{3}x+1{)}^2=1$
$\Rightarrow x^2+4(\sqrt{3}x+1{)}^2=4$
$\Rightarrow x^2+4(3x^2+1+2\sqrt{3}x)=4$
$\Rightarrow 13x^2+8\sqrt{3}x=0$
$x=0$ or $x=-\frac{8\sqrt{3}}{13}$
$y=-\sqrt{3}\left(\frac{8\sqrt{3}}{13}\right)+1=1-\frac{24}{13}=-\frac{11}{13};$
Hence, $B=(\frac{-8\sqrt{3}}{13},-\frac{11}{13})$
$(AB{)}^2={\left(\frac{8\sqrt{3}}{13}\right)}^3+{\left(\frac{24}{13}\right)}^2=\frac{64\times 3+{24}^2}{169}$
$=\frac{192+576}{169}=\frac{768}{169}$
$\Rightarrow AB=\sqrt{\frac{768}{169}}$
$\Rightarrow m+n=768+169=937$