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MCQMediumJEE 2024Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2024 Question with Solution

The area (in square units) of the region enclosed by the ellipse x2+3y2=18x^2 + 3y^2 = 18 in the first quadrant below the line y=xy = x is:

  • A

    3π+34\sqrt{3}\pi + \frac{3}{4}

  • B

    3π\sqrt{3}\pi

  • C

    3π−34\sqrt{3}\pi - \frac{3}{4}

  • D

    3π+1\sqrt{3}\pi + 1

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The ellipse is x2+3y2=18x^2 + 3y^2 = 18 and the required region lies in the first quadrant below the line y=xy = x.

Find: The area of the enclosed region.

From the ellipse,

x218+y26=1\frac{x^2}{18} + \frac{y^2}{6} = 1

so the upper branch is

y=18−x23=18−x231/2.y = \frac{\sqrt{18 - x^2}}{\sqrt{3}} = \frac{\sqrt{18 - x^2}}{3^{1/2}}.

The line y=xy = x meets the ellipse in the first quadrant.

Substitute y=xy = x into x2+3y2=18x^2 + 3y^2 = 18:

x2+3x2=18x^2 + 3x^2 = 18 4x2=184x^2 = 18 x2=92x^2 = \frac{9}{2}

Hence the intersection point is

(32,32).\left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right).

Therefore, the required area is written as

Area=∫032x dx+∫323218−x23 dx.\text{Area} = \int_{0}^{\frac{3}{\sqrt{2}}} x \, dx + \int_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}} \frac{\sqrt{18 - x^2}}{\sqrt{3}} \, dx.

Evaluate the first integral:

∫032x dx=12(x2)032=12(92)=94.\int_{0}^{\frac{3}{\sqrt{2}}} x \, dx = \frac{1}{2}\left(x^2\right)_{0}^{\frac{3}{\sqrt{2}}} = \frac{1}{2}\left(\frac{9}{2}\right) = \frac{9}{4}.

For the second integral, use

∫a2−x2 dx=x2a2−x2+a22sin⁡−1(xa)+C.\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C.

With a=32a = 3\sqrt{2},

∫323218−x2 dx=[x218−x2+9sin⁡−1(x32)]3232.\int_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}} \sqrt{18 - x^2} \, dx = \left[\frac{x}{2}\sqrt{18 - x^2} + 9\sin^{-1}\left(\frac{x}{3\sqrt{2}}\right)\right]_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}}.

So,

Area=94+13[9sin⁡−1(1)−322⋅33−9sin⁡−1(12)].\text{Area} = \frac{9}{4} + \frac{1}{\sqrt{3}}\left[9\sin^{-1}(1) - \frac{3}{2\sqrt{2}} \cdot 3\sqrt{3} - 9\sin^{-1}\left(\frac{1}{2}\right)\right].

Now use

sin⁡−1(1)=π2,sin⁡−1(12)=π6.\sin^{-1}(1) = \frac{\pi}{2}, \qquad \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}.

Thus,

Area=94+13(9π2−934−9π6).\text{Area} = \frac{9}{4} + \frac{1}{\sqrt{3}}\left(\frac{9\pi}{2} - \frac{9\sqrt{3}}{4} - \frac{9\pi}{6}\right).

On simplification,

Area=3π.\text{Area} = \sqrt{3}\pi.

Therefore, the required area is 3π\sqrt{3}\pi, so the correct option is B.

Intersection and Region Setup

Given: The ellipse is x218+y26=1\frac{x^2}{18} + \frac{y^2}{6} = 1 and the line is y=xy = x.

Find: The area in the first quadrant below the line and inside the ellipse.

The ellipse has semi-axes 323\sqrt{2} along the xx-axis and 6\sqrt{6} along the yy-axis. The line y=xy = x divides the first-quadrant part of the ellipse into two pieces.

At first, from x=0x = 0 to x=32x = \frac{3}{\sqrt{2}}, the upper boundary of the required region is the line y=xy = x. After the intersection point, from x=32x = \frac{3}{\sqrt{2}} to x=32x = 3\sqrt{2}, the upper boundary is the ellipse

y=18−x23.y = \sqrt{\frac{18 - x^2}{3}}.

Hence the area must be split into two integrals:

∫032x dx\int_{0}^{\frac{3}{\sqrt{2}}} x \, dx

and

∫323218−x23 dx.\int_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}} \sqrt{\frac{18 - x^2}{3}} \, dx.

Evaluating these gives the same final result:

3π.\sqrt{3}\pi.

Therefore, the correct option is B.

Common mistakes

  • Using the ellipse as the upper boundary over the entire first-quadrant interval is incorrect because from x=0x = 0 to the intersection point, the region is below the line y=xy = x. Split the area at the intersection point and use the correct upper curve on each interval.

  • Solving the intersection incorrectly by substituting y=xy = x into the ellipse and missing that x2+3x2=18x^2 + 3x^2 = 18 gives 4x2=184x^2 = 18. Use this carefully to get x=32x = \frac{3}{\sqrt{2}} in the first quadrant.

  • Writing the ellipse as y=18−x23y = \frac{\sqrt{18 - x^2}}{3} is wrong. Since 3y2=18−x23y^2 = 18 - x^2, we get y=18−x23=18−x23y = \sqrt{\frac{18 - x^2}{3}} = \frac{\sqrt{18 - x^2}}{\sqrt{3}} instead.

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