The area (in sq. units) of the region enclosed by the curve 9x^2 + 4y^2 = 36 is Option 1: 10 Option 2: 8 Option 3: 6 Option 4: 4

Answer: 6. Solution: The equation 9x^2 + 4y^2 = 36 has both x^2 and y^2 terms with different coefficients that are added together. This is an ellipse. The standard form of an ellipse is x^2/a^2 + y^2/b^2 = 1. Dividing the entire equation by 36: 9x^2/36 + 4y^2/36 = 36/36 x^2/4 + y^2/9 = 1 From x^2/4 + y^2/9 = 1: a^2 = 4, so a = 2 b^2 = 9, so b = 3 The area of an ellipse is given by A = ab. A = x 2 x 3 A = 6 Therefore, the area of the region enclosed by the curve is 6 square units.

Calculus

Application of Integrals

Medium

core

The area (in sq. units) of the region enclosed by the curve $9x^2 + 4y^2 = 36$ is

$10\pi$

$8\pi$

$6\pi$

$4\pi$

✅ Correct Option: 3

The equation $9x^2 + 4y^2 = 36$ has both $x^2$ and $y^2$ terms with different coefficients that are added together. This is an ellipse.

The standard form of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ .

Dividing the entire equation by 36:

$\frac{9x^2}{36} + \frac{4y^2}{36} = \frac{36}{36}$

$\frac{x^2}{4} + \frac{y^2}{9} = 1$

From $\frac{x^2}{4} + \frac{y^2}{9} = 1$ :

$a^2 = 4$ , so $a = 2$

$b^2 = 9$ , so $b = 3$

The area of an ellipse is given by $A = \pi ab$ .

$A = \pi \times 2 \times 3$

$A = 6\pi$

Therefore, the area of the region enclosed by the curve is $6\pi$ square units.

2025: 16 May Shift 1

The area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant, is

2026: 21 May Shift 1

The area in the first quadrant bounded by $y = 9x^2$ , $x = 0$ , $y = 1$ and $y = 4$ is :

2022: 7 Aug Shift 2

The area of the region bounded by $f(x) = x|x|$ , x-axis and from $x = -1$ to $x = 1$ is :

2025: 26 May Shift 2

The area (in sq. units) of the region bounded by the curve $x^2 = 250y$ , $y = 0$ and $x = 50$ is

2025: 15 May Shift 2

The area bounded by the curve $y = \log x, y = 0$ and $x = e$ , is

2026: 15 May Shift 1

The area (in sq. units) bounded by the parabola $y^2 = x$ and line $x = 4$ in first quadrant is equal to $A$ , then which of the following statements is/are TRUE ?

A. $A = \dfrac{16}{3}$

B. $A = \displaystyle\int_0^4 \sqrt{x}\, dx$

C. $A = 2\displaystyle\int_0^4 \sqrt{x}\, dx$

D. $A = \displaystyle\int_0^2 y^2\, dy$

Choose the correct answer from the options given below:

The area (in sq. units) of the region enclosed by the curve 9x2+4y2=369x^2 + 4y^2 = 369x2+4y2=36 is

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The area (in sq. units) of the region enclosed by the curve $9x^2 + 4y^2 = 36$ is