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

1 Answer
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 + 20x + 4y - 40y + 100 = 0#

#=> (x^2 + 20x) + (4y - 40y) + 100 = 0#

#=> (x^2 + 20x + 100) + (4y - 40y + 100) + 100 = 0 + 100 + 100#

#=> (x + 10)^2 + 4(y - 5)^2 + 100 = 200#
#=> (x + 10)^2 + 4(y - 5)^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.

#=> ((x + 10)^2 + 4(y - 5)^2)/100 = 100/100#

#=> ((x + 10)^2)/100 + (4(y - 5)^2)/100 = 1#

#=> ((x + 10)^2)/100 + ((y - 5)^2)/25 = 1#

#=> ((x + 10)^2)/10^2 + ((y - 5)^2)/5^2 = 1#

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

#C: (-10, 5)#

#V: (-10 +- 10, 5)#

#f: (-10 +- 5sqrt3, 5)#

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