Question

6.
Find the
area of the
ellipse

Answer 1

Answer:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given equation of ellipse is

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

$\rm :\longmapsto\: \dfrac{ {y}^{2} }{ {b}^{2} } = 1 - \dfrac{ {x}^{2} }{ {a}^{2} }$

$\rm :\longmapsto\: \dfrac{ {y}^{2} }{ {b}^{2} } = \dfrac{ {a}^{2} - {x}^{2} }{ {a}^{2} }$

$\rm :\longmapsto\: {y}^{2} = \dfrac{ {b}^{2} ( {a}^{2} - {x}^{2}) }{ {a}^{2} }$

$\rm :\implies\:y = \dfrac{b}{a} \sqrt{ {a}^{2} - {x}^{2} }$

Curve sketching :-

We know, ellipse is symmetrical in all the four quadrants.

So,

Required area is

$\rm \: \: = \:4 \displaystyle\int_0^a \: y \: dx$

$\rm \: \: = \:4 \displaystyle\int_0^a \: \dfrac{b}{a} \sqrt{ {a}^{2} - {x}^{2} } \: dx$

$\rm \: \: = \:\dfrac{4b}{a} \displaystyle\int_0^a \: \sqrt{ {a}^{2} - {x}^{2} } \: dx$

$\rm \: \: = \:\dfrac{4b}{a} \bigg(\dfrac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + \dfrac{ {a}^{2} }{2} {sin}^{ - 1}\dfrac{x}{a}\bigg)_0^a$

$\rm \: \: = \:\dfrac{4b}{a} \bigg(0 + \dfrac{ {a}^{2} }{2} {sin}^{ - 1}\dfrac{a}{a} - 0 - 0\bigg)$

$\rm \: \: = \:\dfrac{4b}{a} \bigg( \dfrac{ {a}^{2} }{2} {sin}^{ - 1}1\bigg)$

$\rm \: \: = \: \dfrac{4b}{a} \times \dfrac{ {a}^{2} }{2} \times \dfrac{\pi}{2}$

$\rm \: \: = \: \pi \: ab \: square \: units$

Hence,

$\bf :\longmapsto\:Area \: bounded \: by \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1 \: is \: \pi \: ab \: sq. \: units$

Formula Used :-

$\red{ \boxed{ \sf{\displaystyle\int\tt \sqrt{ {a}^{2} - {x}^{2}}dx = \frac{x}{2} \sqrt{ {a}^{2} - {x}^{2}} + \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a} + c }}}$

$\red{ \boxed{ \sf{ {sin}^{ - 1}1 = \frac{\pi}{2} }}}$

$\red{ \boxed{ \sf{ {sin}^{ - 1}0 = 0 }}}$

$\red{ \boxed{ \sf{Area \: w. \: r. \: t. \: x - axis \: = \: \displaystyle\int_{x_1}^{x_2} \:y \: dx }}}$