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Q.

An equilateral triangle is inscribed in the ellipse $x^{2} + 3 y^{2} = 3$ such that one vertex of the triangle is (0, 1) and one altitude of the triangle is along y-axis the length of its side is

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a

$\frac{4 \sqrt{3}}{5}$

b

$\frac{3 \sqrt{3}}{5}$

c

$\frac{6 \sqrt{3}}{5}$

d

$2 \sqrt{3}$

answer is C.

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Detailed Solution

eq $\frac{x^{2}}{3} + \frac{y^{2}}{1} = 1$

Question Image

$Tan 60^{0} = \frac{1 + sinθ}{\sqrt{3} cosθ} \sqrt{3} = \frac{1 + sinθ}{\sqrt{3 cosθ}} 3 cosθ = 1 + sinθ S . B . S 9 (1 - \sin^{2} θ) = {(1 + sinθ)}^{2} 9 (1 - sinθ) = 1 + sinθ 8 = 10 sinθ sinθ = \frac{4}{5} length of side = 2 \sqrt{3} cosθ = 2 \sqrt{3} (\frac{3}{5}) = \frac{6 \sqrt{3}}{5}$

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