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JEE Main 20202 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
2 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2020 · 2 September · Evening Shift · Mathematics · Parabola

The area (in sq. units) of an equilateral triangle inscribed in the...

The area (in sq. units) of an equilateral triangle inscribed in the parabola y2=8x,{y}^{2} = 8x, with one of its vertices on the vertex of this parabola isThe area (in sq. units) of an equilateral triangle inscribed in the parabola y ^ 2 = 8x, with one of its vertices on

Options

  1. A64364\sqrt{3}
  2. B2563256\sqrt{3}
  3. C1923192\sqrt{3}
  4. D1283128\sqrt{3}

Correct answer

C. 1923192\sqrt{3}

Step-by-step solution

The area (in sq. units) of an equilateral triangle inscribed in the parabola y ^ 2 = 8x, with one of its vertices on

y2=8x,a=2{y}^{2} = 8x, a = 2

A≡(2t2,2(2)t)≡(2t2,4t)A ≡ \left(2{t}^{2}, 2 ( 2 ) t\right) ≡ \left(2{t}^{2}, 4t\right)

tan⁡30°=4t2t2=2t=13\tan 30 ° = \dfrac{4t}{2{t}^{2}} = \dfrac{2}{t} = \dfrac{1}{\sqrt{3}}

t=23t = 2\sqrt{3}

Area of δOAB=2⋅δOMA=2⋅12⋅(2t2)(4t)=8t3=8(23)3=1923\delta OAB = 2 \cdot \delta OMA = 2 \cdot \dfrac{1}{2} \cdot \left(2{t}^{2}\right) ( 4t ) = 8{t}^{3} = 8 ( 2\sqrt{3}{)}^{3} = 192\sqrt{3}

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