Question

If the area of the region enclosed by the curve \( ay = x^2 \) and the line \( x + y = 2a \) is \( k a^3 \), then \( k \) is:

• A. \( \frac{2}{9} \)
• B. \( \frac{9}{2} \)
• C. \( \frac{3}{2} \)
• D. \( \frac{2}{3} \)

Hint:

For enclosed area problems, identify the limits of integration from intersection points and subtract the curves correctly.

Answer 1

The Correct Option is B

Step 1: Find the Points of Intersection 
The given equations are: \[ ay = x^2 \quad \text{(Parabola)} \] \[ x + y = 2a \quad \text{(Line)} \] Substituting \( y = 2a - x \) into the parabola equation: \[ a(2a - x) = x^2. \] Rearranging: \[ x^2 + ax - 2a^2 = 0. \] Solving for \( x \) using the quadratic formula: \[ x = \frac{-a \pm \sqrt{a^2 + 8a^2}}{2} = \frac{-a \pm 3a}{2}. \] \[ x = a \quad \text{or} \quad x = -2a. \] Corresponding \( y \) values: \[ \text{For } x = a, \quad y = 2a - a = a. \] \[ \text{For } x = -2a, \quad y = 2a + 2a = 4a. \] 

Step 2: Compute the Area Enclosed 
The enclosed area is given by: \[ A = \int_{-2a}^{a} (2a - x) \, dx - \int_{-2a}^{a} \frac{x^2}{a} \, dx. \] Evaluating the first integral: \[ \int (2a - x) \, dx = 2ax - \frac{x^2}{2}. \] \[ \Bigg[ 2ax - \frac{x^2}{2} \Bigg]_{-2a}^{a}. \] \[ \left[ (2a \cdot a - \frac{a^2}{2}) - (2a(-2a) - \frac{(-2a)^2}{2}) \right]. \] \[ \left[ (2a^2 - \frac{a^2}{2}) - (-4a^2 - 2a^2) \right] = 9a^2. \] Evaluating the second integral: \[ \int \frac{x^2}{a} \, dx = \frac{1}{a} \frac{x^3}{3}. \] \[ \Bigg[ \frac{x^3}{3a} \Bigg]_{-2a}^{a}. \] \[ \left[ \frac{a^3}{3a} - \frac{(-2a)^3}{3a} \right]. \] \[ \left[ \frac{a^2}{3} - (-\frac{8a^2}{3}) \right] = \frac{9a^2}{3} = 3a^2. \] \[ A = 9a^2 - 3a^2 = 6a^2. \] Thus, \[ A = k a^3. \] \[ k = \frac{9}{2}. \]

 Step 3: Conclusion 
Thus, the correct answer is: \[ \mathbf{\frac{9}{2}}. \]