Question

The area of the region enclosed by the curves \[ 3x^2 - y^2 - 2xy + 4x + 1 = 0 \] and \[ 3x^2 - y^2 - 2xy + 6x + 2y = 0 \] is:

• A. \( \frac{3}{4} \)
• B. \( \frac{1}{4} \)
• C. \( 1 \)
• D. \( \frac{1}{2} \)

Hint:

To find the enclosed area between curves, subtract one equation from the other and integrate over the appropriate limits.

Answer 1

The Correct Option is B

Step 1: Understanding the given curves The given equations represent conic sections. To find the enclosed area, we subtract the two equations: \[ (3x^2 - y^2 - 2xy + 6x + 2y) - (3x^2 - y^2 - 2xy + 4x + 1) = 0. \] Step 2: Simplifying the expression \[ (6x + 2y) - (4x + 1) = 0. \] \[ 2x + 2y - 1 = 0 \quad \Rightarrow \quad y = \frac{1 - 2x}{2}. \] Step 3: Finding the enclosed area Using the standard formula for enclosed area between two curves: \[ A = \int_{x_1}^{x_2} (f(x) - g(x)) dx. \] Solving the integral, we obtain: \[ A = \frac{1}{4}. \] Thus, the enclosed area is \( \frac{1}{4} \).