Question

Find the area of small region bounded by the ellipse x²/a² + y²/b²=1 and the line x/a + y/b =1.​

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

Find the area of small region bounded by the curves

$\sf \:  \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1 \: and \: \dfrac{x}{a} + \dfrac{y}{b} = 1$

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\tt \:  \longrightarrow \: \int \: \sqrt{ {a}^{2} - {x}^{2} } dx = \dfrac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + \dfrac{ {a}^{2} }{2} {sin}^{ - 1} \dfrac{x}{a} + c$

$\tt \:  \longrightarrow \: \int \: x \: dx = \dfrac{ {x}^{2} }{2} + c$

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$\large\underline\purple{\bold{Solution :- }}$

Point of intersection of curves :-

$\tt \:  \longrightarrow \: \dfrac{x}{a} + \dfrac{y}{b} = 1$

$\tt\implies \:\dfrac{y}{b} = 1 - \dfrac{x}{a} - - - (1)$

$\tt \:  \longrightarrow \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

☆ On substituting equation (1), we get

$\tt \:  \longrightarrow \: \dfrac{ {x}^{2} }{ {a}^{2} } + {(1 - \dfrac{x}{a}) }^{2} = 1$

$\tt \:  \longrightarrow \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {x}^{2} }{ {a}^{2} } + 1 - \dfrac{2x}{a} = 1$

$\tt \:  \longrightarrow \: \dfrac{ {2x}^{2} }{ {a}^{2} } - \dfrac{2x}{a} = 0$

$\tt\implies \:\dfrac{2x}{a} (\dfrac{x}{a} - 1) = 0$

$\tt\implies \:x \: = 0 \: or \: x \: = a$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf b \\ \\ \sf a & \sf 0 \end{array}} \\ \end{gathered}$

$\tt \:  \longrightarrow \: Required \: Area = \int_0^a(y_{(ellipse)} - y_{(line)}) dx$

$\tt \:  = \int_0^a(\dfrac{b}{a} \sqrt{ {a}^{2} - {x}^{2} }- \dfrac{b}{a} (a - x)) dx$

$\tt \:  =\dfrac{b}{a} [\dfrac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + \dfrac{ {a}^{2} }{2} {sin}^{ - 1}\dfrac{x}{a} - ax + \dfrac{ {x}^{2} }{2} ]_0^a$

$\tt \:  =\dfrac{b}{a}\bigg(0 +\dfrac{ {a}^{2} }{2} {sin}^{ - 1} 1 - {a}^{2} + \dfrac{ {a}^{2} }{2} \bigg)$

$\tt \:  =\dfrac{b}{a}\bigg( \dfrac{ {a}^{2} }{2} \times \dfrac{\pi}{2} - \dfrac{ {a}^{2} }{2} \bigg)$

$\tt \:  =\dfrac{b}{a} \times \dfrac{ {a}^{2} }{2} \bigg(\dfrac{\pi}{2} - 1 \bigg)$

$\tt \: = \dfrac{ab}{2} \bigg( \dfrac{\pi}{2} - 1 \bigg) \: sq. \: units$

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