# How do you find the center, vertices, foci and eccentricity of x^2 + 20x + 4y^2 - 40y + 100 = 0?

Feb 5, 2016

Transform the equation into the standard form

Do this by "completing" the square.
Do this by grouping your $x$s and $y$s, and adding a certain value such that we will end up with a perfect square trinomial. We don't want to ruin the equality, so we need to add the same value on the other side of the equation

${x}^{2} + 20 x + 4 y - 40 y + 100 = 0$

$\implies \left({x}^{2} + 20 x\right) + \left(4 y - 40 y\right) + 100 = 0$

$\implies \left({x}^{2} + 20 x + 100\right) + \left(4 y - 40 y + 100\right) + 100 = 0 + 100 + 100$

$\implies {\left(x + 10\right)}^{2} + 4 {\left(y - 5\right)}^{2} + 100 = 200$
$\implies {\left(x + 10\right)}^{2} + 4 {\left(y - 5\right)}^{2} = 100$

We want the right side of the equation to be equal to 1, so we divide both sides of the equation by 100.

$\implies \frac{{\left(x + 10\right)}^{2} + 4 {\left(y - 5\right)}^{2}}{100} = \frac{100}{100}$

$\implies \frac{{\left(x + 10\right)}^{2}}{100} + \frac{4 {\left(y - 5\right)}^{2}}{100} = 1$

$\implies \frac{{\left(x + 10\right)}^{2}}{100} + \frac{{\left(y - 5\right)}^{2}}{25} = 1$

$\implies \frac{{\left(x + 10\right)}^{2}}{10} ^ 2 + \frac{{\left(y - 5\right)}^{2}}{5} ^ 2 = 1$

Now that the equation is in standard form, we can get the desired properties pretty easily

$C : \left(- 10 , 5\right)$

$V : \left(- 10 \pm 10 , 5\right)$

$f : \left(- 10 \pm 5 \sqrt{3} , 5\right)$

I don't remember how to get the eccentricity, but I'm sure you can already get it from your $a$ and $b$