Question

Find the area bounded by the curve \( y^2 = 4x \) and its latus rectum.

• A. \( \frac{4}{3} \) sq units
• B. \( \frac{8}{3} \) sq units
• C. \( \frac{16}{3} \) sq units
• D. \( \frac{2}{3} \) sq units

Hint:

For the parabola \(y^2 = 4ax\), the latus rectum is \(x=a\). When finding the area bounded by the parabola and its latus rectum, use symmetry about the \(x\)-axis and integrate the positive branch of the curve.

Answer 1

The Correct Option is B

Concept: For the parabola \( y^2 = 4ax \):
• The latus rectum is the line \( x = a \).
• The endpoints of the latus rectum are \( (a,2a) \) and \( (a,-2a) \).
• Area bounded by the parabola and its latus rectum can be obtained by integrating the curve between these limits.

Step 1: Identify the value of \(a\). Given: \[ y^2 = 4x \] Comparing with \( y^2 = 4ax \), we get \[ a = 1 \] Thus, the latus rectum is the line \[ x = 1 \] and its endpoints are \( (1,2) \) and \( (1,-2) \).

Step 2: Express \(y\) in terms of \(x\). \[ y = \pm \sqrt{4x} \] Because the region is symmetric about the \(x\)-axis, \[ \text{Area} = 2\int_{0}^{1} \sqrt{4x}\,dx \]

Step 3: Evaluate the integral. \[ \text{Area} = 2\int_{0}^{1} 2\sqrt{x}\,dx \] \[ = 4\int_{0}^{1} x^{1/2} dx \] \[ = 4\left[\frac{2}{3}x^{3/2}\right]_{0}^{1} \] \[ = 4\left(\frac{2}{3}\right) \] \[ = \frac{8}{3} \]