The area bounded by an ellipse \( \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1 \) and the lines \( x=0,x=ae \) and the \( x \) -axis in the first quadrant, where \( e \) is the eccentricity of the ellipse, is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{ab}{2}\left[\sin^{-1}e+e\sqrt{1-e^2}\right] \)
In the first quadrant the upper half of the ellipse is \( y=\frac{b}{a}\sqrt{a^2-x^2} \), so
\( \displaystyle \text{Area}=\frac{b}{a}\int_{0}^{ae}\sqrt{a^2-x^2}\,dx \)
Using the standard integral
\( \displaystyle \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a} \)
we get
\( \displaystyle \text{Area}=\frac{b}{a}\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_{0}^{ae} \)
At x = ae, \( \sqrt{a^2-a^2e^2}=a\sqrt{1-e^2} \) and \( \sin^{-1}\frac{ae}{a}=\sin^{-1}e \); at x = 0 the bracket is 0:
\( \displaystyle \text{Area}=\frac{b}{a}\left[\frac{ae}{2}\cdot a\sqrt{1-e^2}+\frac{a^2}{2}\sin^{-1}e\right] \)
\( \displaystyle =\frac{b}{a}\cdot\frac{a^2}{2}\left[e\sqrt{1-e^2}+\sin^{-1}e\right] \)
\( \displaystyle =\frac{ab}{2}\left[\sin^{-1}e+e\sqrt{1-e^2}\right] \)