Question

Find the area of the region bounded by the parabola \( y^2 = 4ax \) and its latus rectum.

• A. \( \dfrac{4a^2}{3} \)
• B. \( \dfrac{8a^2}{3} \)
• C. \( \dfrac{16a^2}{3} \)
• D. \( \dfrac{32a^2}{3} \)

Hint:

For regions involving a parabola symmetric about the \(x\)-axis, it is often easier to integrate from \(0\) to the upper limit and multiply the result by \(2\).

Answer 1

The Correct Option is C

Concept: For the parabola \[ y^2 = 4ax \] the latus rectum is the line passing through the focus and perpendicular to the axis of the parabola. Important properties:

• Focus: \( (a,0) \)
• Equation of latus rectum: \( x = a \)
• End points of latus rectum: \( (a,2a) \) and \( (a,-2a) \)

The required area is the region between the parabola \(x=\frac{y^2}{4a}\) and the line \(x=a\) from \(y=-2a\) to \(y=2a\).
Step 1: {Write the area using integration.} \[ A = \int_{-2a}^{2a} \left( a - \frac{y^2}{4a} \right) dy \]
Step 2: {Use symmetry of the curve.} Since the region is symmetric about the \(x\)-axis, \[ A = 2 \int_{0}^{2a} \left( a - \frac{y^2}{4a} \right) dy \]
Step 3: {Evaluate the integral.} \[ A = 2 \left[ ay - \frac{y^3}{12a} \right]_{0}^{2a} \] Substitute the limits: \[ A = 2 \left[ a(2a) - \frac{(2a)^3}{12a} \right] \] \[ A = 2 \left[ 2a^2 - \frac{8a^3}{12a} \right] \] \[ A = 2 \left[ 2a^2 - \frac{2a^2}{3} \right] \] \[ A = 2 \left( \frac{6a^2 - 2a^2}{3} \right) \] \[ A = 2 \left( \frac{4a^2}{3} \right) \] \[ A = \frac{16a^2}{3} \]