The parabola y^2 = 4x divides the area of the circle x^2 + y^2 =… — Solution | JEE Mathematics

MCQMediumJEE 2024Conic Sections (Parabola, Ellipse, Hyperbola)

The parabola $y^2 = 4x$ divides the area of the circle $x^2 + y^2 = 5$ in two parts. The area of the smaller part is:

Answer

Correct answer:A

Standard Method

Given: The curves are $y^2 = 4x$ and $x^2 + y^2 = 5$ .

Find: The area of the smaller part into which the parabola divides the circle.

First find the points of intersection. Put $x = \frac{y^2}{4}$ from the parabola into the circle:

\left(\frac{y^2}{4}\right)^2 + y^2 = 5

\frac{y^4}{16} + y^2 = 5

y^4 + 16y^2 - 80 = 0

Let $z = y^2$ . Then

z^2 + 16z - 80 = 0

Solving,

z = 4 \text{ or } -20

Since $y^2 \ge 0$ , we take $y^2 = 4$ , so $y = \pm 2$ . Hence the curves intersect at $\left(1,2\right)$ and $\left(1,-2\right)$ .

By symmetry about the $x$ -axis, if $A_1$ is the upper half of the smaller region, then the required area is $2A_1$ .

From the given working,

A_1 = \int_{0}^{1} \sqrt{4x} \, dx + \int_{1}^{\sqrt{5}} \sqrt{5 - x^2} \, dx

Evaluating,

A_1 = \frac{4}{3}\left[x^{3/2}\right]_{0}^{1} + \left[\frac{x}{2}\sqrt{5-x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x}{\sqrt{5}}\right)\right]_{1}^{\sqrt{5}}

Substituting the limits,

A_1 = \frac{1}{3} + \frac{5\pi}{4} - \frac{5}{2}\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)

Therefore,

\text{Required area} = 2A_1 = \frac{2}{3} + \frac{5\pi}{2} - 5\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)

Using inverse-trigonometric identity

Use the identity

\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}

So,

\frac{\pi}{2} - \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right)

Hence,

\frac{2}{3} + \frac{5\pi}{2} - 5\sin^{-1}\left(\frac{1}{\sqrt{5}}\right) = \frac{2}{3} + 5\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)

Now,

\cos^{-1}\left(\frac{1}{\sqrt{5}}\right) = \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)

Therefore,

\text{Required area} = \frac{2}{3} + 5\sin^{-1}\left(\frac{2}{\sqrt{5}}\right)

So the correct option is A.

Students may forget to find the intersection points first. Without locating $\left(1,2\right)$ and $\left(1,-2\right)$ , the limits of integration are chosen incorrectly. Always solve the two curve equations together before setting up the area.
A common error is taking only one half of the symmetric region and not doubling it. The figure is symmetric about the $x$ -axis, so after computing the upper half area $A_1$ , the required area is $2A_1$ .
Students often confuse the circle contribution with the whole sector or the wrong segment. The integral from $x=1$ to $x=\sqrt{5}$ corresponds only to the upper circular arc part of the smaller region, not the entire right half of the circle.

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