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Maths Question 18 – JEE-MAIN 2026

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The area of the region bounded by the curves x+3y2=0 and x+4y2=1 is equal to:

Recognize that both equations represent parabolas opening along the negative x-axis.

● Video Walkthrough
Step 1: Identify the curves and intersection points✦ Active

The given curves are x=−3y2 and x=1−4y2. To find their intersection points, we set the expressions for x equal to each other:

−3y2=1−4y24y2−3y2=1y2=1y=±1

The intersection points occur at y=−1 and y=1. When y=±1, x=−3(±1)2=−3. So the intersection points are (−3,1) and (−3,−1).

Step 2: Determine the 'right' and 'left' curves○ Expand

We need to determine which curve is to the right and which is to the left within the interval y∈[−1,1]. Let's test a point, for example, y=0. For x=−3y2, x=0. For x=1−4y2, x=1. Since 1>0, the curve x=1−4y2 is the 'right' curve and x=−3y2 is the 'left' curve.

Step 3: Calculate the area using integration○ Expand

The area A of the region bounded by the curves is given by the integral of the difference between the right and left curves with respect to y from y=−1 to y=1:

A=∫−11((1−4y2)−(−3y2))dyA=∫−11(1−4y2+3y2)dyA=∫−11(1−y2)dy

Now, we evaluate the definite integral:

A=[y−y33]−11A=(1−133)−((−1)−(−1)33)A=(1−13)−(−1−−13)A=(23)−(−1+13)A=23−(−23)A=23+23A=43
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