# How do you find the center, vertices, foci and eccentricity of x^2/100 + y^2/64 = 1?

Nov 11, 2015

The ellipse is in the form

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$

Where

center $= \left(h , k\right)$

focus = $\left(h + c , k\right) , \left(h - c , k\right)$

${c}^{2} = {a}^{2} - {b}^{2}$

${x}^{2} / 100 + {y}^{2} / 64 = 1$

$\implies {\left(x - 0\right)}^{2} / {10}^{2} + {\left(y - 0\right)}^{2} / {8}^{2} = 1$

$\implies$ center $= \left(0 , 0\right)$

${c}^{2} = {a}^{2} - {b}^{2}$

${c}^{2} = 100 - 64$

$\implies {c}^{2} = 36$
$\implies c = 6$

$\implies$ focus $= \left(h + c , k\right) , \left(h - c , k\right)$

$\implies$ focus $= \left(0 + 6 , 0\right) , \left(0 - 6 , 0\right)$

$\implies$ focus $= \left(6 , 0\right) , \left(- 6 , 0\right)$

I don't know how to get the eccentricity.
But I believe it has something to do with the ratio of $a$ and $b$ (or maybe $c$). Sorry