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

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Solution

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

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