
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 ycoordinate 2, I know the major axis is horizontal. I use:
(xh)^2 (yk)^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 + (y2)^2 Thank You(Rofl), Kay
16 b^2

$\displaystyle \frac{(x+3)^{2}}{16} + \frac{(y2)^{2}}{b^{2}} = 1$
Since you are given the point (1, 1) just plug in and solve:
$\displaystyle \frac{(1+3)^{2}}{16} + \frac{(12)^{2}}{b^{2}} = 1$
$\displaystyle \frac{(2)^{2}}{16} + \frac{(3)^{2}}{b^{2}} = 1$
etc. etc.

Hello,
Ok, so far it's correct.
$\displaystyle \frac{(x+3)^2}{16}+\frac{(y2)^2}{b^2}=1$
There is just one thing to do :
(1,1) is on the ellipse... So replace x and y, and find b !