Question

The area of the region common to the parabolas \(4y^2 = 9x\) and \(3x^2 = 16y\) is...

• A. 2 sq. units
• B. 4 sq. units
• C. 8 sq. units
• D. 16 sq. units

Hint:

Find the intersection points, then integrate the upper curve minus the lower curve.

Answer 1

The Correct Option is B

Step 1: Integrate along y
Express both in terms of $y$: $x = \frac{4y^2}{9}$ and $x = \sqrt{\frac{16y}{3}} = \frac{4}{\sqrt3}\sqrt y$.

Step 2: Area
Between $y=0$ and $y=3$: $\int_0^3\left(\frac{4}{\sqrt3}\sqrt y - \frac{4y^2}{9}\right)dy = \frac{4}{\sqrt3}\cdot\frac23\cdot3\sqrt3 - \frac{4}{27}\cdot27 = 8-4 = 4$.

Step 3: Answer
Option (B).

Final Answer:
4 square units. \[ \boxed{\text{(B)}\ 4\ \text{sq. units}} \]