Question

AOB is the positive quadrant of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) in which \(\mathrm{OA}=5, \mathrm{OB}=3\). The area between the arc AB and the chord AB of the ellipse in sq. units is

• A. \(\frac{3}{5}(\pi -2)\)
• B. \(\frac{15}{2}(\pi -2)\)
• C. \(\frac{3}{10}(\pi -2)\)
• D. \(\frac{15}{4}(\pi -2)\)

Hint:

Subtract the area of triangle OAB from the quadrant area of the ellipse.

Answer 1

The Correct Option is D

Step 1: Setup
The chord AB is $y=3-\frac35x$, so Area $=\int_0^5\left(\frac35\sqrt{25-x^2}-\left(3-\frac35x\right)\right)dx$.

Step 2: Evaluate
$\int_0^5\frac35\sqrt{25-x^2}\,dx=\frac35\cdot\frac{25\pi}{4}=\frac{15\pi}{4}$ and $\int_0^5\left(3-\frac35x\right)dx=15-\frac{15}{2}=\frac{15}{2}$.

Step 3: Result
Difference $=\frac{15\pi}{4}-\frac{15}{2}=\frac{15}{4}(\pi-2)$.

Final Answer:
The area equals $\frac{15}{4}(\pi-2)$ square units. \[ \boxed{\text{(D) }\dfrac{15}{4}(\pi-2)} \]