Finding the equation of an ellipse centered at (-1,3) and through the points (1,3) and (-1,4)

Question

I've been taking a computer graphics class and as a review we had some questions about some geometric problems. This one I can not seem to figure out though it seems that it should be straight forward.

I know that the standard equation of an ellipse is

x^2/a^2 + y^2/b^2 = 1

From this we shift to its stated center and the equation becomes

(x+1)^2/a^2 + (y-3)^2/b^2 =1

But from here I am at a loss for where to go from here...

Draw a picture of your ellipse. Remember that in your equation, $a$ is the length from the center of the ellipse to its rightmost point and $b$ is the length from the centre to its lowest point. You can see what those should be. — Ragib Zaman, Sep 21 '13 at 14:45
Plug the coordinates of the given points into your equation. — Michael Hoppe, Sep 21 '13 at 14:46

amWhy · Accepted Answer · 2013-09-21T14:56:29.733

2

Substitute into your second equation (corrected with $b^2$ in the second denominator) the $x, y$ values for each of the points on the ellipse $(1, 3), (-1, 4)$, and you'll obtain two equations in two unknowns, from which you can solve for $a^2$ and $b^2$.

That is, evaluate $$\dfrac{(x+1)^2}{a^2} + \dfrac{(y-3)^2}{b^2} =1$$

at (1) $(1, 3)$ and again at (2) $(-1, 4)$. That will give you two equations in $a^2, b^2$, for which you can solve simultaneously.

$$\dfrac{(1 + 1)^2}{a^2} + \dfrac{(3-3)^2}{b^2} = 1 \implies \dfrac 4{a^2} = 1 \implies a^2 = 4\tag{1}$$

$$\dfrac{(-1 + 1)^2}{a^2} + \dfrac{(4-3)^2}{b^2} = 1 \implies \dfrac 1{b^2} = 1 \implies b^2 = 1\tag{2}$$

Now, use the general equation with your new-found values for $a^2, b^2$.

edited Sep 21 '13 at 14:56

answered Sep 21 '13 at 14:49

amWhy

209,954

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from the given three conditions, how can we assume that the axes of the ellipse are parallel to the Cartesian axes? – lab bhattacharjee Sep 21 '13 at 14:58
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@labbhattacharjee There is only one ellipse with a given center passing through two specified points, so any ellipse you find satisfying those conditions is the only one. – Ragib Zaman Sep 21 '13 at 15:01
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@RagibZaman, how can we establish the uniqueness? – lab bhattacharjee Sep 21 '13 at 15:02
@labbhattacharjee You're making this much more difficult than it needs to be! – amWhy Sep 21 '13 at 15:03
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God created all ellipses with their axes parallel to x and y axes. It was the mathematicians who twisted things. – Maesumi Sep 21 '13 at 15:12
Hahaha, I like that, @Maesumi! – amWhy Sep 21 '13 at 15:13
@amWhy, why don't you supply direct answer if you have one? – lab bhattacharjee Sep 21 '13 at 15:16
@amWhy: Another TU +1 – Amzoti Sep 22 '13 at 00:27

Finding the equation of an ellipse centered at (-1,3) and through the points (1,3) and (-1,4)

1 Answers1