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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Ellipse

NTA Abhyas 2020 · Mathematics · Ellipse

An equilateral triangle is inscribed in the ellipse whose equation...

An equilateral triangle is inscribed in the ellipse whose equation is x2+4y2=4.x^{2} + 4y^{2} = 4 . One vertex of the triangle is (0,1)\left(0,1\right) and one altitude is contained in the~yy-axis. If the length of each side is k3k\sqrt{3} units, then kk isAn equilateral triangle is inscribed in the ellipse whose equation is x^ 2 + 4y^ 2 = 4 . One vertex of the triangle is

Options

  1. A1613\dfrac{16}{13}
  2. B813\dfrac{8}{13}
  3. C1316\dfrac{13}{16}
  4. D138\dfrac{13}{8}

Correct answer

A. 1613\dfrac{16}{13}

Step-by-step solution

An equilateral triangle is inscribed in the ellipse whose equation is x^ 2 + 4y^ 2 = 4 . One vertex of the triangle is

x24+y21=1\dfrac{x^{2}}{4} + \dfrac{y^{2}}{1} = 1
Equation of ABAB is& y=3x+1y = \sqrt{3}x + 1

x2+4(3x2+23x+1)=4x^{2} + 4\left(3x^{2} + 2\sqrt{3}x + 1\right) = 4
⇒13x2+83x=0⇒x=−8313⇒ 13x^{2} + 8\sqrt{3}x = 0 ⇒ x = \dfrac{- 8\sqrt{3}}{13}
BC=16313BC = \dfrac{16\sqrt{3}}{13}& units

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