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

NTA Abhyas 2020 · Mathematics · Ellipse

A trapezium is formed by the pair of tangents of parabola P: y = x^...

A trapezium is formed by the pair of tangents of parabola P:y=x24+1P: y = \dfrac{x^{2}}{4} + 1 drawn from the centre of the ellipse E:x24+y2=14E: \dfrac{x^{2}}{4} + y^{2} = \dfrac{1}{4}, tangent at the vertex of PP and the tangent at end point of the minor axis of EE. The area (in sq. units) of the trapezium isA trapezium is formed by the pair of tangents of parabola P: y = x^ 2 4 + 1 drawn from the centre of the ellipse E: x^

Options

  1. A34\dfrac{3}{4}
  2. B32\dfrac{3}{2}
  3. C316\dfrac{3}{16}
  4. D38\dfrac{3}{8}

Correct answer

A. 34\dfrac{3}{4}

Step-by-step solution

A trapezium is formed by the pair of tangents of parabola P: y = x^ 2 4 + 1 drawn from the centre of the ellipse E: x^

The equation of the tangents from the centre of the ellipse to the parabola are y=±xy = \pm x

The equation of the tangent at the vertex PP of the parabola& is y=1y = 1

The equation of the tangent at the end point of the minor axis of the ellipse is y=12y = \dfrac{1}{2}

Hence, the area of trapezium ABCD=12⋅12(1+2)=34ABCD= \dfrac{1}{2} ⋅ \dfrac{1}{2} \left(1 + 2\right) = \dfrac{3}{4}& sq. units

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