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Question Number 41488 by behi83417@gmail.com last updated on 08/Aug/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18

	$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{(\frac{a^{2}}{b^{2}})} = 1 \frac{x^{2}}{(\frac{a^{2}}{b^{2}})} + \frac{y^{2}}{a^{2}} = 1$ $a^{2} > \frac{a^{2}}{b^{2}} (a s s u m e d)$ $s o l v e$ $x^{2} + b^{2} y^{2} = a^{2} p u t t i n g t h e v a l u e {o f x}^{2} i n 2 n d e q n$ $b^{2} x^{2} + y^{2} = a^{2}$ $b^{2} (a^{2} - b^{2} y^{2}) + y^{2} = a^{2}$ $a^{2} b^{2} + y^{2} (1 - b^{4}) = a^{2}$ $y^{2} (1 - b^{4}) = a^{2} (1 - b^{2})$ $y^{2} = \frac{a^{2}}{1 + b^{2}} s o y = \pm \frac{a}{\sqrt{1 + b^{2}}}$ $x^{2} + b^{2} y^{2} = a^{2}$ $x^{2} + b^{2} (\frac{a^{2}}{1 + b^{2}}) = a^{2}$ $x^{2} = a^{2} - \frac{a^{2} b^{2}}{1 + b^{2}}$ $x^{2} = \frac{a^{2} + a^{2} b^{2} - a^{2} b^{2}}{1 + b^{2}}$ $x = \pm \frac{a}{\sqrt{1 + b^{2}}} a n d y = \frac{a}{\sqrt{1 + b^{2}}}$ $c o n t d$
	Commented by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18
	$=4[∫_0 ^(a/(√(1+b^2 ))) (1/(b.))(√(a^2 −x^2 )) dx+∫_(a/(√(1+b^2 ))) ^(a/b) b.(√((a^2 /b^2 )−x^2 )) dx] =4[(1/b)∣(x/2)(√(a^2 −x^2 )) +(a^2 /2)sin^(−1) ((x/a))∣_0 ^(a/(√(1+b^2 ))) ]+b[∣(x/2)(√((a^2 /b^2 )−x^2 )) +((a^2 /b^2 )/2)sin^(−1) ((x/(a/b)))∣_(a/(√(1+b^2 ))) ^(a/b) ] =4[(1/b){(a/(2(√(1+b^2 )))).(√(a^2 −(a^2 /(1+b^2 )))) +(a^2 /2)sin^(−1) ((1/(√(1+b^2 ))))}+ b. [(0+(a^2 /(2b^2 )).(Π/2))−((a/(2(√(1+b^2 ))))(√((a^2 /b^2 )−_ (a^2 /(1+b^2 )))) +(a^2 /(2b^2 ))sin^(−1) (b/(√(1+b^2 ))))$
	$= 4 [\int_{0}^{\frac{a}{\sqrt{1 + b^{2}}}} \frac{1}{b .} \sqrt{a^{2} - x^{2}} d x + \int_{\frac{a}{\sqrt{1 + b^{2}}}}^{\frac{a}{b}} b . \sqrt{\frac{a^{2}}{b^{2}} - x^{2}} d x]$ $= 4 [\frac{1}{b} ∣ \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} {s i n}^{- 1} (\frac{x}{a}) {\overset{\frac{a}{\sqrt{1 + b^{2}}}}{∣}}_{0}] + b [∣ \frac{x}{2} \sqrt{\frac{a^{2}}{b^{2}} - x^{2}} + \frac{\frac{a^{2}}{b^{2}}}{2} {s i n}^{- 1} (\frac{x}{\frac{a}{b}}) {\overset{\frac{a}{b}}{∣}}_{\frac{a}{\sqrt{1 + b^{2}}}}]$ $= 4 [\frac{1}{b} {\frac{a}{2 \sqrt{1 + b^{2}}} . \sqrt{a^{2} - \frac{a^{2}}{1 + b^{2}}} + \frac{a^{2}}{2} {s i n}^{- 1} (\frac{1}{\sqrt{1 + b^{2}}})} +$ $b . [(0 + \frac{a^{2}}{2 b^{2}} . \frac{Π}{2}) - (\frac{a}{2 \sqrt{1 + b^{2}}} \sqrt{\frac{a^{2}}{b^{2}} -_{} \frac{a^{2}}{1 + b^{2}}} + \frac{a^{2}}{2 b^{2}} {s i n}^{- 1} \frac{b}{\sqrt{1 + b^{2}}})$
	Commented by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18

	$i s h a l l c o m p l e t e t h e p r o b l e m i n c o p y t h e n p o s t . .$ $p l d w a i t . . .$
	Commented by behi83417@gmail.com last updated on 08/Aug/18

	$d e a r t a n m a y s i r! t h a n k y o u f o r$ $s o h a r d w o r k .$
	Answered by MJS last updated on 08/Aug/18

	$(1) x^{2} + b^{2} y^{2} - a^{2} = 0$ $(2) b^{2} x^{2} + y^{2} - a^{2} = 0 \Rightarrow y_{1} = \sqrt{a^{2} - b^{2} x^{2}}$ $(1 - 2) (1 - b^{2}) x^{2} + (b^{2} - 1) y^{2} = 0$ $\Rightarrow x^{2} = y^{2}$ $in (1) y^{2} + b^{2} y - a^{2} = 0 \Rightarrow y = \pm \frac{a}{\sqrt{1 + b^{2}}} \Rightarrow x = \pm \frac{a}{\sqrt{1 + b^{2}}}$ $area = square + 4 \times bow$ $square = {(\frac{2 a}{\sqrt{1 + b^{2}}})}^{2} = \frac{4 a^{2}}{1 + b^{2}}$ $4 \times bow = 2 \times 4 \times \underset{\frac{a}{\sqrt{1 + b^{2}}}}{\int^{\frac{a}{b}}} y_{1} d x = 2 π \frac{a^{2}}{b} - \frac{4 a^{2}}{1 + b^{2}} - \frac{4 a^{2}}{b} \arctan b$ $area = \frac{2 a^{2}}{b} (π - 2 \arctan b)$ $(will post the work to \int y_{1} d x later)$
	Commented by behi83417@gmail.com last updated on 08/Aug/18

	$n i c e & s m a r t . t h a n k s i n a d v a n c e s i r .$
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