Question

An equilateral triangle is inscribed in an ellipse whose equation is $x^2+4 y^2=4$. 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

$ \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 $
$x^2+4\left(3 x^2+1+2 \sqrt{3} x\right)=4 \Rightarrow 13 x^2+8 \sqrt{3} x=0$
$x=0 \text { or } x=-\frac{8 \sqrt{3}}{13} \text { in }(2)$
$y=-\sqrt{3}\left(\frac{8 \sqrt{3}}{13}\right)+1=1-\frac{24}{13}=-\frac{11}{13} ; \text { Hence } B=\left(-\frac{8 \sqrt{3}}{13},-\frac{11}{13}\right)$
$(A B)^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 A B=\sqrt{\frac{768}{169}} $
$\Rightarrow m +n=768+169=937$