Question

Q.22. Find the area enclosed by the ellipse x²/16+y²/9=1 ?​

Answer 1

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

Given curve is

$\rm \: \dfrac{ {x}^{2} }{16} + \dfrac{ {y}^{2} }{9} = 1$

Its an equation of ellipse having centre (0, 0) and intersects the x - axis at (4, 0) and (- 4, 0) and intersects the y - axis at (0, 3) and (0, - 3)

Also, Ellipse is symmetrical in all the four quadrants.

$\rm \: \dfrac{ {y}^{2} }{9} = 1 - \dfrac{ {x}^{2} }{16}$

$\rm \: \dfrac{ {y}^{2} }{9} = \dfrac{16 - {x}^{2} }{16}$

$\rm \: {y}^{2} = 9\bigg(\dfrac{16 - {x}^{2} }{16}\bigg)$

$\rm\implies \:y = \dfrac{3}{4} \sqrt{16 - {x}^{2} }$

So, required area is 4 times the area bounded by ellipse in first quadrant with respect to x - axis between the lines x = 0 and x = 4.

$\rm \: Area = 4 \times \displaystyle\int_{0}^{4}\rm \: ydx$

$\rm \: = 4 \times \dfrac{3}{4} \displaystyle\int_{0}^{4}\rm \sqrt{16 - {x}^{2} } \: dx$

$\rm \: = 3 \times \displaystyle\int_{0}^{4}\rm \sqrt{ {4}^{2} - {x}^{2} } \: dx$

$\rm \: = 3\bigg( \frac{x}{2} \sqrt{ {4}^{2} - {x}^{2} } + \frac{ {4}^{2} }{2} {sin}^{ - 1} \frac{x}{4} \bigg)_{0}^{4}$

$\rm \: = 3\bigg( 0 + \frac{ {4}^{2} }{2} {sin}^{ - 1} \frac{4}{4} - 0 - 0 \bigg)$

$\rm \: = 3\bigg( 8 {sin}^{ - 1}1 \bigg)$

$\rm \: = 24 \times \dfrac{\pi}{2}$

$\rm \: = 12\pi \: square \: units$

Hence,

$\rm\implies \:\boxed{ \rm{ \:Area\bf \: = 12\pi \: square \: units \: \: }}$

$\rule{190pt}{2pt}$

Short Cut Trick :-

$\boxed{ \rm{ \:Area \: bounded \: by \: ellipse \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1 \: is \: \pi \: ab \: sq. \: units \: }}$

Formula Used :-

$\boxed{ \rm{\int \sqrt{ {a}^{2}-{x}^{2}}dx=\frac{x}{2} \sqrt{ {a}^{2}-{x}^{2} }+\frac{ {a}^{2}}{2}{sin}^{-1} \frac{x}{a}+c}}$