question_answer

Find the area of the smaller region bounded by the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] and the line \[\frac{x}{a}+\frac{y}{b}=1.\]

Answer:

Given equations of ellipse and the straight line are \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]                           ?(i) and       \[\frac{x}{a}+\frac{y}{b}=1\]                             ?(ii) Ellipse with Eq. (i) has vertices \[(\pm \,\,a,\,\,0)\]and \[(0,\pm \,\,b,)\] and centre (0, 0). While the line with Eq. (ii) has x-intercept  a and y-intercept b.                                So, line passes through the points (a, 0) and (0, b). \[\therefore \] Graph of the above region is given below                    Clearly, points of intersection are A(a, 0) and  B(0, b) \[\therefore \] Required area \[=\int_{0}^{a}{[{{y}_{(ellipse)}}-{{y}_{(line)}}]}\,dx\]         ?(iii) Now, equation of ellipse is \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\Rightarrow \]   \[\frac{{{y}^{2}}}{{{b}^{2}}}=1-\frac{{{x}^{2}}}{{{a}^{2}}}=\frac{{{a}^{2}}-{{x}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \]   \[{{y}^{2}}=\frac{{{b}^{2}}}{{{a}^{2}}}({{a}^{2}}-{{x}^{2}})\] \[\Rightarrow \]   \[y=\frac{b}{a}\sqrt{{{a}^{2}}-{{x}^{2}}}\]                          ?(iv) and equation of line is \[\frac{x}{a}+\frac{y}{b}=1\] \[\Rightarrow \]   \[\frac{y}{b}=1-\frac{x}{a}=\frac{a-x}{a}\] \[\Rightarrow \]   \[y=\frac{b}{a}(a-x)\]                            ?(v) Hence, from Eqs. (iii), (iv) and (v), Required area \[=\int_{0}^{a}{\left[ \frac{b}{a}\sqrt{{{a}^{2}}-{{x}^{2}}}-\frac{b}{a}(a-x) \right]\,dx}\]             \[=\frac{b}{a}\int_{0}^{a}{\sqrt{{{a}^{2}}-{{x}^{2}}}dx-\frac{b}{a}\int_{0}^{a}{(a-x)}\,dx}\] We know that, \[\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a}+C\] \[\therefore \]Area \[=\frac{b}{a}\left[ \frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a} \right]_{0}^{a}-\frac{b}{a}\left[ ax-\frac{{{x}^{2}}}{2} \right]_{0}^{a}\] \[=\frac{b}{a}\left[ \frac{{{a}^{2}}}{2}{{\sin }^{-1}}1 \right]_{0}^{a}-\frac{b}{a}\left[ {{a}^{2}}-\frac{{{a}^{2}}}{2} \right]\] \[=\frac{b{{a}^{2}}}{2a}{{\sin }^{-1}}\left( \sin \frac{\pi }{2} \right)-\frac{b}{a}\left( \frac{{{a}^{2}}}{2} \right)\] \[\left[ \because \,\,1=\sin \frac{\pi }{2}\Rightarrow {{\sin }^{-1}}1={{\sin }^{-1}}\left( \sin \frac{\pi }{2} \right) \right]\] \[=\left( \frac{ba}{2}\times \frac{\pi }{2} \right)-\frac{ab}{2}\] \[=\frac{\pi ab}{4}-\frac{ab}{2}=\left( \frac{\pi }{4}-\frac{1}{2} \right)\] ab sq units