# Ellipse Equation

• Apr 28th 2008, 11:27 AM
K.Chopin
Ellipse Equation
My instructor writes this on our review sheet, "An ellipse with vertices (-7,2) and (1,2) passes through (-1,-1). Write the equation."

Since the vertices have y-coordinate 2, I know the major axis is horizontal. I use:
(x-h)^2 (y-k)^2
a^2 + b^2

So I find the center (-3,2)
I also find a= 4

So, how do I find b^2?(Speechless) My instructor did not provide the minor axis or "c."

So far I have:

(x+3)^2 + (y-2)^2 Thank You(Rofl), Kay
16 b^2
• Apr 28th 2008, 11:34 AM
o_O
$\displaystyle \frac{(x+3)^{2}}{16} + \frac{(y-2)^{2}}{b^{2}} = 1$

Since you are given the point (-1, -1) just plug in and solve:
$\displaystyle \frac{(-1+3)^{2}}{16} + \frac{(-1-2)^{2}}{b^{2}} = 1$
$\displaystyle \frac{(2)^{2}}{16} + \frac{(-3)^{2}}{b^{2}} = 1$

etc. etc.
• Apr 28th 2008, 11:36 AM
Moo
Hello,

Ok, so far it's correct.

$\displaystyle \frac{(x+3)^2}{16}+\frac{(y-2)^2}{b^2}=1$

There is just one thing to do :

(-1,-1) is on the ellipse... So replace x and y, and find b !