foci (0, +/- 4/3), contains (4/5,1)

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- Apr 29th 2012, 04:27 PMsluggerbrothfind equation of ellipse????
foci (0, +/- 4/3), contains (4/5,1)

- Apr 29th 2012, 04:55 PMskeeterRe: find equation of ellipse????
centered at the origin (why?)

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\displaystyle c = \frac{4}{3}$

$\displaystyle b^2 = a^2 + c^2$

same drill ... sub in $\displaystyle x$ and $\displaystyle y$, solve for $\displaystyle a^2$ and $\displaystyle b^2$ - Apr 29th 2012, 05:02 PMsluggerbrothRe: find equation of ellipse????
algera gets really, really messy??? or am i doing something wrong?????

- Apr 29th 2012, 05:26 PMskeeterRe: find equation of ellipse????
yes, this one is messy ...

$\displaystyle \frac{(4/5)^2}{a^2} + \frac{1^2}{b^2} = 1$

$\displaystyle b^2 = a^2 + \frac{16}{9}$

$\displaystyle \frac{16}{25a^2} + \frac{1}{b^2} = 1$

$\displaystyle \frac{16}{25a^2} + \frac{1}{a^2 + \frac{16}{9}} = 1$

$\displaystyle \frac{16}{25a^2} + \frac{9}{9a^2 + 16} = 1$

$\displaystyle 16(9a^2+16) + 9(25a^2} = 25a^2(9a^2+16)$

$\displaystyle 144a^2 + 256 + 225a^2 = 225a^4 + 400a^2$

$\displaystyle 0 = 225a^4 -31a^2 - 256$

$\displaystyle 0 = (225a^2-256)(a^2 + 1)$

$\displaystyle a^2 = \frac{256}{225}$

sometimes you just have to grind it out