Ellipse

vanbeersj

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Aug 6, 2008
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I have to find the equation of the ellipse with foci (1,4) and (1,0) that passes through (4,4) by use of the definition.

So the definition of the ellipse the text provides is sqrt((x-c)^2 +y^2) + sqrt((x+c)^2 +y^2) = 2a
I enter to get sqrt((x-1)^2 +(y-4)^2) + sqrt((x-1)^2 + (y-0)^2)
After I expand and simplify I'm left with

a = x^2-2x+y^2-4y+9
but I'm not sure I did it right because I'm suppose to be able to sketch the curve with the points and I can't do that because I don't know where the centre is located. Please help.
 
ok then by using the definition I have the following:

sqrt[(x-1)^2 + (y-4)^2] + [sqrt(x-1)^2 + (y-0)^2] = 4

If I simplify this and reduce, I get x^2-2x+y^2-4y +1 = 0

How do I go from here to isolate x so that I can sketch this curve?
 
vanbeersj said:
ok then by using the definition I have the following:

sqrt[(x-1)^2 + (y-4)^2] + [sqrt(x-1)^2 + (y-0)^2] = 4

If I simplify this and reduce, I get x^2-2x+y^2-4y +1 = 0

How do I go from here to isolate x so that I can sketch this curve?

The foci are at (1, 4) and (1, 0)....another helper told you that the center of the ellipse is halfway between the foci.

You need to START by finding the center....you haven't done that yet. Until you find the correct coordinates for the center, you're going to have trouble.

And, I'm not sure why you think you need to isolate x to be able to sketch the curve. When you know the center of an ellipse, its orientation (vertical or horizontal), and the lengths of the major and minor axis, sketching the ellipse should be relatively easy.
 
If the centre is between the foci, then I would have a centre point of (1,2)
My text only walks me through how to do this if the centre is at the origin (0,0), it states that to find the equation of an ellipse with its centre at a point other than the origin to use the definition of the ellipse; which is sqrt[(x-c)^2 + y^2] + sqrt[(x+c)^2 +y^2 = 2a, I understand the foci is c, and that a = major axis, which is the distance from the vertex to the centre. But with the foci and the centre I'm not sure how to find the major axis in order to start the equation, I mistakenly thought it was 4 as outlined in the prior work above. How can I find a?
 
The "definition" that you posted applies to an ellipse centered at the origin with a horizontal major axis, where the sum of the distances from the foci to the given point is 2a.

This "definition" is generally used to derive the basic form x^2/a^2 + y^2/b^2 = 1; I have not seen it used for anything else.

You seem to think that you need to use this definition to derive something for a shifted vertical ellipse; I would take a different approach.

Generally, we're taught how to work with ellipses centered at the origin, and then we apply transformations to shift them.

Have you learned about transformations?

EG:

Replacing x with (x - h) shifts the graph h units horizontally (to the right when h is positive; to the left when h is negative).

Replacing y with (y - k) shifts the graph k units vertically (up when k is positive; down when k is negative).

Therefore, we can find the equation of the given ellipse centered at the origin, and then apply the shifts to write its equation.

We know that the major axis is vertical. Here is the formula for such an ellipse, when the center is at the origin.

\(\displaystyle \frac{x^2}{b^2} \;+\; \frac{y^2}{a^2} \;=\; 1\)

(b is half the length of the major axis, and a is half the length of the minor axis)

The foci are at (0, c) and (0, -c), where c^2 = a^2 - b^2.

Shifting this ellipse one unit to the right and two units up gives us the following shifted form.

\(\displaystyle \frac{(x - 1)^2}{b^2} \;+\; \frac{(y - 2)^2}{a^2} \;=\; 1\)

Therefore, to get the equation for the ellipse in this exercise, we need to find the values of a and b.

Here's what we know about this ellipse, when it's centered at (0, 0).

The foci are at (0, 2) and (0, -2), so the value of c is 2.

\(\displaystyle c^2 \;=\; a^2 - b^2 \quad \Longrightarrow \quad b^2 \;=\; a^2 - 4\)

The point (3, 2) is on this ellipse, so we have the following equation.

\(\displaystyle \frac{9}{a^2 - 4} \;+\; \frac{4}{a^2} \;=\; 1\)

This equation is sufficient to find the value of a^2. Once we know that, we can find the value of b^2. Then, we have everything we need to (1) write the equation of the shifted ellipse, (2) calculate the semi-axis lengths, and (3) determine the endpoint coordinates of both the major and minor axes, to make a sketch.

I get a major axis of 8 units, and a minor axis of 6.9 units (rounded). And you?

Let us know, if you need more help. 8-)

Cheers ~ Mark

PS: I might fool around with that "definition" after dinner, to see how it could be used, but I'm thinking that the method above is way easier.
 
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