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 $\sqrt{m/n}$, where $m$ and $n$ are relatively prime positive integers. Find the value of $(m+n)$.

Answer 1

Answer: 937

$\frac{x^2}{4}+y^2=1$ .....(1)
equation of the side $AB$
$y-1=\sqrt{3}(x-0)\Rightarrow y=\sqrt{3}x+1$....(2)
solving (1) and (2)
$\frac{x^2}{4}+(\sqrt{3}x+1{)}^2=1\Rightarrow x^2+4(\sqrt{3}x+1{)}^2=4$
$x^2+4\left(3x^2+1+2\sqrt{3}x\right)=4\Rightarrow 13x^2+8\sqrt{3}x=0$
$x=0\text{ or }x=-\frac{8\sqrt{3}}{13}\text{ in }(2)$
$y=-\sqrt{3}\left(\frac{8\sqrt{3}}{13}\right)+1=1-\frac{24}{13}=-\frac{11}{13};\text{ Hence }B=\left(-\frac{8\sqrt{3}}{13},-\frac{11}{13}\right)$
$(AB{)}^2={\left(\frac{8\sqrt{3}}{13}\right)}^3+{\left(\frac{24}{13}\right)}^2=\frac{64\cdot 3+24^2}{169}=\frac{192+576}{169}=\frac{768}{169}\Rightarrow AB=\sqrt{\frac{768}{169}}$
$\Rightarrow m+n=768+169=937$