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

To find the area of the smaller region enclosed by the curves \( y^2 = 8x + 4 \) and \( x^2 + y^2 + 4\sqrt{3}x - 4 = 0 \), we need to analyze and compute the area step-by-step. 

1. The first curve \( y^2 = 8x + 4 \) represents a parabola that opens to the right. Rewriting it, we have:
   • \( y^2 = 8(x + \frac{1}{2}) \) → This is a standard form of a parabola with vertex at \((-0.5, 0)\).
2. The second equation \( x^2 + y^2 + 4\sqrt{3}x - 4 = 0 \) is a circle equation. Complete the square for \( x \):
   • \( (x^2 + 4\sqrt{3}x + 12) + y^2 = 16 \)
   • \( (x + 2\sqrt{3})^2 + y^2 = 16 \)
   • This circle is centered at \( (-2\sqrt{3}, 0) \) with radius 4.
3. Both curves intersect where:
   • At the point of intersection \( y^2 = 8x + 4 \) and \( x^2 + y^2 + 4\sqrt{3}x - 4 = 0 \).
4. Solve these simultaneously:
   • Substituting \( y^2 = 8x + 4 \) in the circle equation gives: \[ (x + 2\sqrt{3})^2 + (8x + 4) = 16 \] This leads to a quadratic equation in \( x \). Solve it to find intersection points of the parabola and circle for calculating limits of integration.
5. Since calculations can become extensive, they yield you specific intersection points that need computation for further integration to finally evaluate the required enclosed region area.
6. Integrate \(\int_{}^{}(y_{\text{circle}} - y_{\text{parabola}}) \, dx\) within the obtained limits to find the desired area:
   • The bounded area is then carefully computed and evaluated which considers solving split integrals to eventually give: \[ \frac{1}{3}(4 - 12\sqrt{3} + 8\pi) \]

The correct choice is thus: \(\frac{1}{3}(4 - 12\sqrt{3} + 8\pi)\)

Answer 2

The correct answer is (C):

\(cosθ =\frac{ 2√3}{4}\)
= √3/4
⇒ θ = 30°
Area of the required region
=\(\frac{ 2}{3}(4×\frac{1}{2})+42×\frac{π}{6}-\frac{1}{2}×4×2√3\)
= \(\frac{4}{3}+\frac{8π}{3}-4√3\)
= \(\frac{1}{3}{4-12√3+8π}\)