# If I am just trying to create an ellipse with the 4 vertices and center, can I just plug the numbers into the equation of an ellipse without worrying about the foci?

Do I need foci to calculate an ellipse?
I have been trying to find an answer, but where I look does not tell me why I need foci if I have all 4 vertices and the center. If I am just trying to create an ellipse with the 4 vertices and center, can I just plug the numbers into the equation of an ellipse without worrying about the foci? I am trying to understand how the foci come into play, as they don't appear in the actual equation of an ellipse. However, I want my ellipse to be correct. I am trying to take a circle, and scale the y axis only, elongating the circle to create the ellipse that still passes through the 4 points, 2 now scaled. It is a vertical ellipse.
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Kash Osborn
Step 1
The general equation of an ellipse is a quadratic expression $a{x}^{2}+2bxy+c{y}^{2}+dx+ey+f=0$ where the foci do not appear explicitly.
Step 2
In the case of an axis-parallel ellipse, the cross term xy does not exist, and you can use the simplified form $\frac{\left(x-{x}_{c}{\right)}^{2}}{{a}^{2}}+\frac{\left(y-{y}_{c}{\right)}^{2}}{{b}^{2}}=1$ where the center and axis lengths are explicit.
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Wyatt Weeks
Explanation:
You have 2a,2b from the given points. If you want to calculate an ellipse from relation $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$ which has axes parallel to coordinate axes that is taken implied for your exercise, you can use $y=b\sqrt{1-{x}^{2}/{a}^{2}}$ to find y for each x.
(Foci location, latus rectum, eccentricity) are related parameters that is not required for above calculation).