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  4. An equilateral triangle is inscribed in the ellipse whose equation is x2+4y2=4. One vertex of the triangle is (0,1) and one altitude is contained in the y -axis. If the length of each side is k√3 units, then k is

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Q. An equilateral triangle is inscribed in the ellipse whose equation is $x^{2}+4y^{2}=4.$ One vertex of the triangle is $\left(0,1\right)$ and one altitude is contained in the $y$ -axis. If the length of each side is $k\sqrt{3}$ units, then $k$ is

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Solution:

Solution
$\frac{x^{2}}{4}+\frac{y^{2}}{1}=1$
Equation of $AB$ is $y=\sqrt{3}x+1$
$x^{2}+4\left(3 x^{2} + 2 \sqrt{3} x + 1\right)=4$
$\Rightarrow 13x^{2}+8\sqrt{3}x=0\Rightarrow x=\frac{- 8 \sqrt{3}}{13}$
$BC=\frac{16 \sqrt{3}}{13}$ units