Question

The area of region bounded by the curve \[ y^2=4ax \] and the straight line \[ x=2a,\qquad a>0 \] in the first quadrant is:

• A. \(\dfrac{8a^2}{3}\ \text{sq. units}\)
• B. \(\dfrac{8\sqrt2\,a^2}{3}\ \text{sq. units}\)
• C. \(\dfrac{32a^2}{3}\ \text{sq. units}\)
• D. \(\dfrac{64a^2}{3}\ \text{sq. units}\)

Hint:

For area between curves: \[ \text{Area}=\int(\text{Right curve}-\text{Left curve})\,dy \] when integrating with respect to \(y\).

Answer 1

The Correct Option is B

Step 1: Write parabola in terms of \(y\) Given: \[ y^2=4ax \] \[ x=\frac{y^2}{4a} \] The line is: \[ x=2a \]
Step 2: Find points of intersection Substitute: \[ x=2a \] in parabola: \[ y^2=4a(2a) \] \[ y^2=8a^2 \] \[ y=2\sqrt2\,a \] Since region lies in first quadrant: \[ 0\leq y\leq2\sqrt2\,a \]
Step 3: Form area integral Area bounded by line and parabola: \[ A=\int_0^{2\sqrt2 a} \left( 2a-\frac{y^2}{4a} \right)dy \]
Step 4: Evaluate the integral \[ A= \left[ 2ay-\frac{y^3}{12a} \right]_0^{2\sqrt2 a} \] Substitute upper limit: \[ A= 2a(2\sqrt2 a) - \frac{(2\sqrt2 a)^3}{12a} \] \[ = 4\sqrt2 a^2 - \frac{16\sqrt2 a^3}{12a} \] \[ = 4\sqrt2 a^2 - \frac{4\sqrt2 a^2}{3} \] \[ = \frac{12\sqrt2 a^2-4\sqrt2 a^2}{3} \] \[ = \frac{8\sqrt2 a^2}{3} \] Therefore: \[ \boxed{ \frac{8\sqrt2\,a^2}{3} } \] Option analysis:
• Option (A): Incorrect
• Option (B): Correct
• Option (C): Incorrect
• Option (D): Incorrect Hence: \[ \boxed{\text{(B)}} \]