Question

Find the area bounded by the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]  and the ordinates x = ae and x = 0, where b² = a² (1 − e²) and e < 1.

 

 

Answer 1

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\text{ represents a parabola , symmetrical about both the axis }. \]
\[\text{ It cuts }x \text{ axis at }A\left( a, 0 \right) \text{ and }A'\left( - a, 0 \right)\]
\[\text{ It cuts }y \text{ axis at }B\left( 0, b \right)\text{ and }B'\left( 0, - b \right)\]
\[x = \text{ ae is a line parallel to }y\text{ axis }\]
\[\text{ Consider a vertical strip of length }= \left| y \right| \text{ and width = dx , in the first quadrant }\]
\[\text{ Area of approximating rectangle in first quadrant }= \left| y \right| dx\]
\[\text{ Approximating rectangle moves from }x = 0\text{ to }x = \text{ ae }\]
\[\text{ Area of the shaded region }= 2 \text{ area in the first quadrant }\]
\[ \Rightarrow A = 2 \int_0^{ae} \left| y \right| dx\]
\[ \Rightarrow A = 2 \int_0^{ae} y dx ..............\left\{ As, y > 0 , \left| y \right| = y \right\}\]
\[ \Rightarrow A = 2 \int_0^{ae} \frac{b}{a}\sqrt{a^2 - x^2}dx ..................\left\{ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \Rightarrow y = \frac{b}{a}\sqrt{a^2 - x^2} \right\}\]
\[ \Rightarrow A = \frac{2b}{a} \int_0^{ae} \sqrt{a^2 - x^2}dx \]
\[ \Rightarrow A = \frac{2b}{a} \left[ \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \frac{x}{a} \right]_0^{ae} \]
\[ \Rightarrow A = \frac{2b}{a}\left[ \frac{1}{2}ae\sqrt{a^2 - a^2 e^2} + \frac{1}{2} a^2 \sin^{- 1} \frac{ae}{a} - 0 \right]\]
\[ \Rightarrow A = \frac{b}{a}\left[ a^2 e\sqrt{1 - e^2} + \frac{1}{2} a^2 \sin^{- 1} e \right]\]
\[ \Rightarrow A = \frac{b}{a} a^2 \left[ e\sqrt{1 - e^2} + \frac{1}{2} \sin^{- 1} e \right]\]
\[ \Rightarrow A = ab\left[ e\sqrt{1 - e^2} + \frac{1}{2} \sin^{- 1} e \right]\text{ sq . units }\]
\[ \therefore\text{ Required area of the ellipse bound by }x = 0 \text{ and }x = \text{ ae is }= ab\left[ e\sqrt{1 - e^2} + \frac{1}{2} \sin^{- 1} e \right]\text{ sq . units }\]