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Question Number 135695 by liberty last updated on 15/Mar/21

W h a t i s t h e e q u a t i o n o f

e l l i p s e w i t h c e n t e r (1, - 2)

e x c e n t r i c i t y \frac{1}{\sqrt{3}}, p a s s i n g

t h r o u g h (2, \frac{2 \sqrt{2} - 2}{3})

Answered by mr W last updated on 15/Mar/21

t h e r e i s n o u n i q u e s o l u t i o n! i t

d e p e n t s o n t h e d i r e c t i o n o f t h e

m a j o r a x i s! s i n c e t h e m a j o r a x i s

i s n o t g i v e n, t h e r e a r e i n f i n i t e

p o s s i b i l i t i e s .

a s s u m e t h e m a j o r a x i s i s p a r a l l e l

t o x - a x i s, t h e n t h e e q n . o f e l l i p s e

i s \frac{{(x - 1)}^{2}}{a^{2}} + \frac{{(y + 2)}^{2}}{b^{2}} = 1

e = \sqrt{1 - {(\frac{b}{a})}^{2}} = \frac{1}{\sqrt{3}}

\Rightarrow \frac{b^{2}}{a^{2}} = \frac{2}{3}

\frac{{(2 - 1)}^{2}}{a^{2}} + \frac{{(\frac{2 \sqrt{2} - 2}{3} + 2)}^{2}}{b^{2}} = 1

\frac{1}{a^{2}} + \frac{8 {(\sqrt{2} + 1)}^{2}}{9 b^{2}} = 1

\frac{1}{a^{2}} + \frac{8 {(\sqrt{2} + 1)}^{2}}{9 \times \frac{2}{3} a^{2}} = 1

a^{2} = 1 + \frac{4 {(\sqrt{2} + 1)}^{2}}{3} = 5 + \frac{8 \sqrt{2}}{3}

b^{2} = \frac{2}{3} (5 + \frac{8 \sqrt{2}}{3})

\Rightarrow {(x - 1)}^{2} + \frac{3 {(y + 2)}^{2}}{2} = 5 + \frac{8 \sqrt{2}}{3}

s e e t h e r e d o n e i n d i a g r a m . b u t t h i s

i s o n l y o n e o f i n f i n i t e p o s s i b i l i t i e s .

Commented by mr W last updated on 15/Mar/21

Commented by liberty last updated on 15/Mar/21

a g r e e s i r

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