The area of the smaller region bounded by the ellipse x^2/9 + y^2/4 = 1 and the line x/3 + y/2 = 1 is : Option 1: 2/3( - 2) Option 2: 2/3( + 2) Option 3: 3/2( - 2) Option 4: 3/2( + 2)

Answer: 3/2( - 2). Solution: Area of quarter ellipse = ( ab)/4 = ( * 3 * 2)/4 = (3)/2. Area of triangle = 1/2(3)(2) = 3. Smaller region = (3)/2 - 3 = 3/2( - 2).

Calculus

Application of Integrals

Medium

The area of the smaller region bounded by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the line $\frac{x}{3} + \frac{y}{2} = 1$ is :

$\frac{2}{3}(\pi - 2)$

$\frac{2}{3}(\pi + 2)$

$\frac{3}{2}(\pi - 2)$

$\frac{3}{2}(\pi + 2)$

✅ Correct Option: 3

Area of quarter ellipse $= \frac{\pi ab}{4} = \frac{\pi \cdot 3 \cdot 2}{4} = \frac{3\pi}{2}$ . Area of triangle $= \frac{1}{2}(3)(2) = 3$ . Smaller region $= \frac{3\pi}{2} - 3 = \frac{3}{2}(\pi - 2)$ .

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The area of the smaller region bounded by the ellipse x29+y24=1\frac{x^2}{9} + \frac{y^2}{4} = 19x2​+4y2​=1 and the line x3+y2=1\frac{x}{3} + \frac{y}{2} = 13x​+2y​=1 is :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The area of the smaller region bounded by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the line $\frac{x}{3} + \frac{y}{2} = 1$ is :