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JEE Mains · Maths · STD 12 - 8. Application and integration

The area of the region bounded by the curves \(x+3y^2=0\) and \(x+4y^2=1\) is equal to:

  1. A \(\dfrac{1}{3}\)
  2. B \(\dfrac{2}{3}\)
  3. C \(\dfrac{4}{3}\)
  4. D \(\dfrac{5}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\dfrac{4}{3}\)

Step-by-step Solution

Detailed explanation

The equations of the given curves are \(x = -3y^2\) and \(x = 1 - 4y^2\). To find the points of intersection, equate the values of \(x\): \(-3y^2 = 1 - 4y^2\) \(y^2 = 1 \Rightarrow y = \pm 1\) The region is bounded between \(y = -1\) and \(y = 1\). In this interval,…