Question

If $AOB$ is the positive quadrant of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in which $OA=a,OB=b.$ Then area enclosed between acr $AB$ and chord $AB$ of the ellipse is

• A. $\frac{ab}{2}(\pi -4)$ sq. unit
• B. $ab(\pi -2)$ sq. unit
• C. $\frac{ab}{4}(\pi -2)$ sq. unit
• D. $\frac{ab}{4}(\pi -4)$ sq. unit

Answer 1

The correct option is C $\frac{ab}{4}(\pi -2)$ sq. unit


Area of ellipse $=\pi ab$ sq. unit
Then area of ellipse in first quadrant $=\frac{1}{4}\pi ab$ sq. unit
Now area of triangle $OAB=\frac{1}{2}ab$ sq. unit
Hence, the required area is $\frac{1}{4}\pi ab-\frac{1}{2}ab=\frac{ab}{4}(\pi -2)$ sq. unit