AP DSC School Assistant Math Model Test 1

Concept:The region bounded by the parabola $y^2 = 4ax$ and its latus rectum is symmetric about the $x$-axis. The latus rectum is the vertical line $x = a$ through the focus $(a,0)$. The required area is twice the area under the upper branch from $x=0$ to $x=a$, computed using definite integration.Explanation:Step 1: The equation of the parabola is $y^2 = 4ax$. Solving for $y$ gives $y = \pm 2\sqrt{a}\sqrt{x}$. The upper half is $y = 2\sqrt{a}\sqrt{x}$.Step 2: The latus rectum is the line $x = a$. The area bounded by the parabola and this line lies between $x = 0$ and $x = a$.Step 3: Due to symmetry about the $x$-axis, the total area $A$ is twice the area under the upper branch:$$A = 2 \int_{0}^{a} 2\sqrt{a}\sqrt{x} \, dx = 4\sqrt{a} \int_{0}^{a} x^{1/2} \, dx.$$Step 4: Evaluate the integral:$$\int_{0}^{a} x^{1/2} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{a} = \frac{2}{3} a^{3/2}.$$Thus,$$A = 4\sqrt{a} \cdot \frac{2}{3} a^{3/2} = \frac{8}{3} \cdot a^{1/2} \cdot a^{3/2} = \frac{8a^{2}}{3}.$$Answer:Option B: $\frac{8a^{2}}{3}$.