# Equation of an Ellipse

• Apr 21st 2009, 06:49 PM
IAmLegendToo
Equation of an Ellipse
Find an equation of an ellipse satisfying the given conditions.

Foci: (-1,6) and (-1,0); Length of major axis: 10

Need more of a check on my answer. If it's wrong, then I'll need some explanations on how to do the problem. If right, then that's all I really need.

The center is obviously the midpoint between the 2 Foci, so it's (-1,3) and it's vertical.

So it's equation is:

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

A is half the length of the major axis (10) so a would = 5.
A squared = 25

C is x in the foci, so it's 1.
C squared = 1

So to find b^2, I need to plug values into:

C^2 = a^2 - b^2
1=25-b^2
b^2 = 24

So, assuming all of the above is correct, the final equation should be:

(x+1)^2/24 + (y-3)^2/25 = 1

Correct? Thanks.
• Apr 21st 2009, 07:08 PM
Soroban
Hello, IAmLegendToo!

Quote:

Find an equation of an ellipse satisfying the given conditions.

Foci: (-1,6) and (-1,0); Length of major axis: 10

Need more of a check on my answer.

The center is obviously the midpoint between the 2 Foci, so it's (-1,3) and it's vertical.

So it's equation is: .$\displaystyle \frac{(x+1)^2}{b^2} + \frac{(y-3)^2}{a^2} \;=\; 1$

$\displaystyle a$ is half the length of the major axis (10) so: .$\displaystyle a = 5 \quad\Rightarrow\quad a^2 = 25$ . all correct!

C is x in the foci, so it's 1 . . . . no

$\displaystyle c$ is the distance from the center to a focus: .$\displaystyle c = 3$

Give it another try . . .

• Apr 21st 2009, 07:11 PM
IAmLegendToo
Ahh, that sounds more correct. I don't know why I thought of it as simply X in a foci. Hmm.

So then it's:

C=3
C^2 = 9
9 = 25 - b^2
b^2 = 16
b= 4

So, then equation would be:

(x+1)^2/16 + (y-3)^2/25 = 1

Correct now?