# Find the equation of ellipse given vertices and focus Check

1. Mar 27, 2005

### aisha

Find the equation of ellipse given vertices and focus Check plz

Hi the question is find the equation of the following ellipse, given vertices at (8,3) and (-4,3) and one focus at (6,3)

Well I drew a digram with the 3 points

First I found the midpoint of the given vertices to get the center of the ellipse I got (h,k) to be (2,3) then I found the distance between the center and the vertex a=6

the only thing I wasnt sure about was how to find b, but this is what I did I found the distance between the two vertices and got 12, I think this is also the legnth of the major axis therefore 2b=12 so b=6

My final equation for the ellipse is

$$\frac {(x-2)^2} {36} + \frac {(y-3)^2} {36} =1$$ Help me out is this correct?

2. Mar 27, 2005

### dextercioby

The equation that u've written describes a circle and not an ellipse...

Do'em all again...

Daniel.

3. Mar 27, 2005

### Data

think about your method of finding $b$ and see if you can figure out something wrong with it (hint: using your method, could you ever have $b \neq a$? Does $b$ really represent the length of the major axis?)

Once you've thought about that, see if you can deduce where the other focus is (remember, ellipses have two of them!)

4. Mar 27, 2005

### aisha

I can probably find where the other focus but how will this help me? How do I find b? Is the rest of the equation correct?

Oh b does not represent the length of the major axis um it represent the length of the minor axis? 2b?

I DONT HAVE A CLUE on HOW TO FIND B!!! HELPPPPPPP

Last edited: Mar 27, 2005
5. Mar 27, 2005

### Data

The rest is fine. $b$ is the length of the semi-minor axis, so it's given by $b^2 = a^2 - c^2$ where $c$ is half the distance between the focii.

Do you see why $b$ has this value?

Last edited: Mar 27, 2005
6. Mar 27, 2005

### aisha

See my diagram doesnt have semi-major axis on it thats why I dont know how to find b, um I got the distance between the foci to be 8 half of this is 4 when I plugged a and -c I got
$$b^2=6^2 =4^2$$ b = sqrt (20) ? I dont really see why b has this value if this is correct...

7. Mar 27, 2005

### Data

Sorry, as has been corrected now I meant semi-minor (here $a$ is actually the semi-major axis, silly terminology). Your answer for $b$ is correct nonetheless.

Do you remember what at ellipse is?

Given two points $P_0 = (x_0, y_0)$ and $P_1 = (x_1, y_1)$, an ellipse with semi-major axis $a$ and focii $P_0$ and $P_1$ is the set of points $P = (x,y)$ such that the sum of the distances from $P$ to $P_0$ and from $P$ to $P_1$ is $2a$.

Can you sketch this shape? Once you do this, find $c$ using the method I described above, and try to see if you can figure out why $b$ is what it is, by constructing triangles with base $c$ and hypotenuse $a$ within the shape.

8. Mar 27, 2005

### aisha

Ok yes I see the triangle so finally the final equation is

$$\frac {(x-2)^2} {36} + \frac {(y-3)^2} {20} =1$$

9. Mar 27, 2005

### Data

Indeed

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