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

NTA Abhyas 2020 · Mathematics · Parabola

Let points A _ 1 , A _ 2 ~and& A _ 3 lie on the parabola y^ 2 = 8x...

Let points A1,A2{A}_{1}, {A}_{2}~and& A3{A}_{3} lie on the parabola y2=8x.y^{2} = 8x . If △A1A2A3△ {A}_{1}{A}_{2}{A}_{3} is an equilateral triangle and normals at points A1,A2{A}_{1}, {A}_{2}~and& A3{A}_{3} on this parabola meet at the point \left(h, 0\right),thenthevalueof, then the value ofh$& isLet points A _ 1 , A _ 2 ~and& A _ 3 lie on the parabola y^ 2 = 8x . If △ A _ 1 A _ 2 A _ 3 is an equilateral triangle

Options

  1. A2424
  2. B2626
  3. C3838
  4. D2828

Correct answer

D. 2828

Step-by-step solution

Let points A _ 1 , A _ 2 ~and& A _ 3 lie on the parabola y^ 2 = 8x . If △ A _ 1 A _ 2 A _ 3 is an equilateral triangle
One of the normal from (h,0)\left(h, 0\right) on the parabola y2=8xy^{2} = 8x is y=0y = 0
hence, assume A3=(0,0)<br/>Let,{A}_{3} = \left(0, 0\right)<br />Let,{A}_{1} = \left(2{t}_{1}^{2}, 4{t}_{1}\right)<br/>and<br />and{A}_{2} = \left(2{t}_{1}^{2}, - 4{t}_{1}\right)
Then,theslopeofOA1the slope of{OA}_{1} is \dfrac{4{t}_{1}}{2{t}_{1}^{2}} = \dfrac{2}{{t}_{1}}<br/><br />⇒ tan {30}^{o} = \dfrac{2}{{t}_{1}}⇒t1=⇒ {t}_{1} =2\sqrt{3}<br/>Theequationofthenormalat<br />The equation of the normal at{A}_{1}isisy = - {t}_{1}x + 4{t}_{1} + 2{t}_{1}^{3}<br/><br />⇒ y = - 232\sqrt{3} x + 83+8\sqrt{3} ++48\sqrt{3}<br/>Putting(h,0)<br />Putting \left(h, 0\right) in the above equation, we get,
0=−230 = - 2\sqrt{3}h + 56\sqrt{3}⇒h=28.⇒ h = 28.
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