Question

The area bounded by the parabolas \(y^2 = 4a(x + a)\) and \(y^2 = -4a(x - a)\) is

• A. \(\frac{16}{3}a^2\)
• B. \(\frac{8}{3}a^2\)
• C. \(\frac{4}{3}a^2\)
• D. None of these

Hint:

For symmetric regions, integrate with respect to y or x as convenient.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The two parabolas are symmetric about the y-axis. Find intersection points and integrate.

Step 2: Detailed Explanation:
First parabola: \(y^2 = 4a(x+a)\) opens right, vertex at \((-a,0)\). Second: \(y^2 = -4a(x-a)\) opens left, vertex at \((a,0)\). They intersect when \(4a(x+a) = -4a(x-a) \Rightarrow x+a = -x+a \Rightarrow 2x=0 \Rightarrow x=0\). Then \(y^2 = 4a(a)=4a^2 \Rightarrow y = \pm 2a\). So symmetric about x-axis. Area = \(2 \int_{y=0}^{2a} (x_R - x_L) dy\). From first: \(x = \frac{y^2}{4a} - a\). From second: \(x = a - \frac{y^2}{4a}\). So \(x_R - x_L = (a - y^2/4a) - (y^2/4a - a) = 2a - \frac{y^2}{2a}\). Area = \(2 \int_0^{2a} (2a - \frac{y^2}{2a}) dy = 2 [2ay - \frac{y^3}{6a}]_0^{2a} = 2[4a^2 - \frac{8a^3}{6a}] = 2[4a^2 - \frac{4a^2}{3}] = 2[\frac{12a^2 - 4a^2}{3}] = 2 \times \frac{8a^2}{3} = \frac{16a^2}{3}\).

Step 3: Final Answer:
\(\frac{16}{3}a^2\), which corresponds to option (A).