An equilateral triangle is inscribed in an ellipse whose equation is x2 + 4y2= 4.If one vertex of the triangle is(0, 1), then the length of each side is

We have x24+y2=1-----1Equation of the side AB: y−1=3(x−0)⇒ y=3x+1----2Solving (1) and (2)x24+(3x+1)2=1⇒x2+43x2+1+23x=4⇒13x2+83x=0⇒x=0 or x=−8313 Putting x=−8313 in (2), we get y=−38313+1=1−2413=−1113 Hence B≡−8313,−1113∴ C≡8313,−1113∴ BC=16313

An equilateral triangle is inscribed in an ellipse whose equation is x2 + 4y2= 4.If one vertex of the triangle is(0, 1), then the length of each side is

We have x24+y2=1-----1Equation of the side AB: y−1=3(x−0)⇒ y=3x+1----2Solving (1) and (2)x24+(3x+1)2=1⇒x2+43x2+1+23x=4⇒13x2+83x=0⇒x=0 or x=−8313 Putting x=−8313 in (2), we get y=−38313+1=1−2413=−1113 Hence B≡−8313,−1113∴ C≡8313,−1113∴ BC=16313

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An equilateral triangle is inscribed in an ellipse whose equation is x² + 4y²= 4.If one vertex of the triangle is
(0, 1), then the length of each side is

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$\frac{16 \sqrt{3}}{13}$

$\frac{48 \sqrt{3}}{13}$

$\frac{8 \sqrt{3}}{13}$

$\frac{24 \sqrt{3}}{13}$

answer is C.

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Detailed Solution

$We have \frac{x^{2}}{4} + y^{2} = 1 - - - - - (1)$

Equation of the side AB:

$\begin{array}{l} y - 1 = \sqrt{3} (x - 0) \\ \Rightarrow y = \sqrt{3} x + 1 - - - - (2) \end{array}$

Solving (1) and (2)
$\begin{array}{l} \frac{x^{2}}{4} + (\sqrt{3} x + 1)^{2} = 1 \\ \Rightarrow x^{2} + 4 (3 x^{2} + 1 + 2 \sqrt{3} x) = 4 \\ \Rightarrow 13 x^{2} + 8 \sqrt{3} x = 0 \\ \Rightarrow x = 0 or x = - \frac{8 \sqrt{3}}{13} \end{array}$

$Putting x = - \frac{8 \sqrt{3}}{13} in (2), we get$

$y = - \sqrt{3} (\frac{8 \sqrt{3}}{13}) + 1 = 1 - \frac{24}{13} = - \frac{11}{13}$

$\begin{array}{l} Hence B \equiv (- \frac{8 \sqrt{3}}{13}, - \frac{11}{13}) \\ ∴ C \equiv (\frac{8 \sqrt{3}}{13}, - \frac{11}{13}) \\ ∴ BC = \frac{16 \sqrt{3}}{13} \end{array}$