Mathematics - Area Under Curves Question with Solution | TestHub

The given equations are $\frac{x^2}{9}+\frac{y^2}{4}=1$ and $\frac{x}{3}+\frac{y}{2}=1$.

The line intersects the ellipse at $(3,0)$ and $(0,2)$, which are the vertices of the ellipse in the first quadrant.

The required area is the region in the first quadrant bounded by the ellipse and the line.

Area of the ellipse in the first quadrant is $\frac{1}{4}\pi ab=\frac{1}{4}\pi (3)(2)=\frac{3\pi }{2}$.

Area of the triangle formed by the line and the coordinate axes is $\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 3\times 2=3$.

The required area is the difference between these two areas:

Area $=\frac{3\pi }{2}-3=\frac{3}{2}(\pi -2)$ sq. units.

Answer: $\frac{3}{2}(\pi -2)$ sq. units