To find the area bounded by the parabolas \(y^2 = 4a(x + a)\) and \(y^2 = -4a(x - a)\), let's first understand the orientation and location of these parabolas on the coordinate plane.
- The equation \(y^2 = 4a(x + a)\) represents a rightward opening parabola with vertex at \((-a, 0)\).
- The equation \(y^2 = -4a(x - a)\) represents a leftward opening parabola with vertex at \((a, 0)\).
Next, let's find their points of intersection to determine the limits of integration.
- Set the equations equal to each other to find the intersection points: \(y^2 = 4a(x + a) = -4a(x - a)\)
- This simplifies to: \(x + a = -(x - a)\)
- Simplifying gives: \(x + a = -x + a\) \(2x = 0\)
- Thus, \(x = 0\)
- Substitute \(x = 0\) into \(y^2 = 4a(x + a)\) to find \(y\): \(y^2 = 4a^2 \Rightarrow y = \pm 2a\)
The points of intersection are \((0, 2a)\) and \((0, -2a)\).
Now, let's calculate the area between these curves from \(y = -2a\) to \(y = 2a\).
The equations for \(x\) in terms of \(y\) from the parabolas are:
- From \(y^2 = 4a(x + a)\): \(x = \frac{y^2}{4a} - a\)
- From \(y^2 = -4a(x - a)\): \(x = a - \frac{y^2}{4a}\)
The bounded area can be calculated as:
- \( \text{Area} = \int_{-2a}^{2a} \left[(a - \frac{y^2}{4a}) - (\frac{y^2}{4a} - a)\right] \, dy\)
- This simplifies to: \(\text{Area} = \int_{-2a}^{2a} \left[2a - \frac{y^2}{2a}\right] \, dy\)
- Integrating: \(\text{Area} = \left[2ay - \frac{y^3}{6a}\right]_{-2a}^{2a}\)
- Substitute \(y = 2a\) and \(y = -2a\): \(\left[2a(2a) - \frac{(2a)^3}{6a}\right] - \left[2a(-2a) - \frac{(-2a)^3}{6a}\right]\)
- \(= 4a^2 - \frac{8a^3}{6a} - (-4a^2) + \frac{8a^3}{6a}\)
- Simplifying gives: \(\text{Area} = 8a^2 - \frac{8a^2}{3}\)
- Final result: \(\text{Area} = \frac{16}{3}a^2\)
Thus, the area bounded by the parabolas is \(\frac{16}{3}a^2\), confirming that the correct answer is: \(\frac{16}{3}a^2\).