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