Question

The area of the smaller region enclosed by the curves y² = 8x + 4 and
x²+y²+4√3x-4=0
is equal to

• A. \(\frac{1}{3}(2-12√3+8π)\)
• B. \(\frac{1}{3}(2-12√3+6π)\)
• C. \(\frac{1}{3}(4-12√3+8π)\)
• D. \(\frac{1}{3}(4-12√3+6π)\)

Answer 1

The Correct Option is C

Let's solve the problem to find the area of the smaller region enclosed by the given curves: \(y^2 = 8x + 4\) and \(x^2 + y^2 + 4\sqrt{3}x - 4 = 0\).

1. First, understand the given equations:
   • The equation \(y^2 = 8x + 4\) represents a parabola. For simplicity, rewrite it as \(y^2 = 8(x + \frac{1}{2})\).
   • The equation \(x^2 + y^2 + 4\sqrt{3}x - 4 = 0\) can be rearranged to represent a circle. Complete the square for the \(x\)-terms:
   • This represents a circle with center \((-2\sqrt{3}, 0)\) and radius 4. 
2. To find the area of intersection, we need to find the points of intersection of these curves:
   • Substitute \(y^2 = 8x + 4\) into the circle equation:
   • Solve this quadratic equation to find the \(x\)-coordinates of intersection points.
3. To find the area of the enclosed region, calculate the definite integrals of the functions between intersection points.
4. Finally, compute the area and determine the correct option:
   • This matches the correct answer from the options: \(\frac{1}{3}(4-12\sqrt{3}+8\pi)\).

Thus, the area of the smaller region enclosed by the curves is \(\frac{1}{3}(4-12\sqrt{3}+8\pi)\).