Let y = p(x) be the parabola passing through the points (–1, 0), (0, 1) and (1, 0). If the area of the region \(\{x,y):(x+1)^2+(y-1)^2≤1,y≤p(x)\}\) is A, then 12(π-4A) is equal to__.
Correct Answer: 16
Solution and Explanation
Given: The parabola is \( x^2 = -4a(y - 1) \), and it passes through the points \( (-1, 1) \) and \( (1, 0) \).
- Step 1: Find the value of \( a \):
\( x^2 = -4a(y - 1) \implies 1 = -4a(-1) \implies a = \frac{1}{4} \).
- Thus, the equation of the parabola becomes:
\( x^2 = -(y - 1) \).
- Step 2: Area covered by the parabola:
The area under the parabola is given by:
\( \int_{-1}^{0} (1 - x^2) \, dx \).
- Integrate:
\( \int_{-1}^{0} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{0} \).
- Substitute the limits:
\( \left[ (0 - 0) \right] - \left[ (-1 + \frac{1}{3}) \right] = \frac{2}{3} \).
- Step 3: Required area:
The required area is the area of the sector minus the area of the square minus the area under the parabola:
\( \text{Required Area} = \frac{\pi}{4} - 1 - \frac{2}{3} \).
- Step 4: Simplify the calculation:
\( \text{Required Area} = \frac{\pi}{4} - \frac{1}{3} \).
- Multiply through by 12 to simplify:
\( \text{Required Area} = 12 \times \frac{\pi - 4}{12} = \frac{\pi - 4}{3} \).
Final Answer: The required area is \( 16 \).
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