Question

The area enclosed by \[ x^2 + 4y^2 \le 4,\qquad y \le |x| - 1,\qquad y \ge 1 - |x| \] is equal to:

• A. \(4\sin^{-1}\!\left(\dfrac{3}{5}\right) + \dfrac{6}{5}\)
• B. \(\sin^{-1}\!\left(\dfrac{3}{5}\right) - \dfrac{6}{5}\)
• C. \(4\sin^{-1}\!\left(\dfrac{3}{5}\right) + \dfrac{12}{5}\)
• D. \(4\sin^{-1}\!\left(\dfrac{3}{5}\right) - \dfrac{6}{5}\)

Hint:

For area problems with symmetry:
Use symmetry to reduce computation
Convert ellipse to standard form
Carefully identify intersection points

Answer 1

The Correct Option is D

Step 1: Rewrite the ellipse: \[ x^2 + 4y^2 = 4 \;\Rightarrow\; \frac{x^2}{4} + y^2 = 1 \] This is an ellipse symmetric about both axes.
Step 2: The lines \[ y = |x| - 1 \text{and} y = 1 - |x| \] form a symmetric diamond-shaped region.
Step 3: By symmetry, compute area in the first quadrant and multiply by 4. In the first quadrant: \[ y = 1 - x \text{intersects ellipse } \frac{x^2}{4}+y^2=1 \] Substitute: \[ \frac{x^2}{4} + (1-x)^2 = 1 \Rightarrow 5x^2 - 8x + 0 = 0 \Rightarrow x = \frac{3}{5} \]
Step 4: Area in one quadrant: \[ \int_{0}^{3/5} \sqrt{1-\frac{x^2}{4}}\,dx \;-\; \int_{0}^{3/5} (1-x)\,dx \] Evaluating: \[ = \sin^{-1}\!\left(\frac{3}{5}\right) - \frac{3}{10} \]
Step 5: Multiply by 4: \[ \text{Required Area} = 4\sin^{-1}\!\left(\frac{3}{5}\right) - \frac{6}{5} \]