# Thread: Graphing Off-Center Conics

1. ## Graphing Off-Center Conics

I'm a bit lost at the moment

I was unfortunately absent on the day of graphing off-center conics in our Algebra 2 class. I've got the notes, but the teacher refused to teach me the subject before or after class. I'm going in tomorrow before school to talk with her, but in the meantime, I could use some help with this.

For instance;

How do I write an equation and graph the ellipse with foci at (4,2) and vertices at (4,4) and (4,-8) ?

I understand standard form, but I don't understand how to add the center points.

Thanks in advance to everyone!

2. Hello ScottStedman

Welcome to Math Help Forum!
Originally Posted by ScottStedman
I'm a bit lost at the moment

I was unfortunately absent on the day of graphing off-center conics in our Algebra 2 class. I've got the notes, but the teacher refused to teach me the subject before or after class. I'm going in tomorrow before school to talk with her, but in the meantime, I could use some help with this.

For instance;

How do I write an equation and graph the ellipse with foci at (4,2) and vertices at (4,4) and (4,-8) ?

I understand standard form, but I don't understand how to add the center points.

Thanks in advance to everyone!
If the centre of the ellipse is at $\displaystyle (h, k)$ then the equation is:

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

where $\displaystyle 2a$ and $\displaystyle 2b$ are, as usual, the lengths of the axes. This is simply because $\displaystyle (x-h) = 0$ when $\displaystyle x =h$ and $\displaystyle (y-k) = 0$ when $\displaystyle y = k$, so the centre has shifted from $\displaystyle (0,0)$ to $\displaystyle (h,k)$.

In the example you've given, the centre is at the mid-point of the line joining $\displaystyle (4,4)$ to $\displaystyle (4,-8)$; i.e. at $\displaystyle (4,-2)$. So $\displaystyle h = 4, k =-2$.

The length of the vertical axis is $\displaystyle 12$ (the distance between these vertices), so $\displaystyle b=6$.

The foci always lie on the major axis, so since one focus is at $\displaystyle (4,2)$, the major axis is vertical, and the foci are at a distance $\displaystyle be$ from the centre. So $\displaystyle be = 4$. And, when the major axis is vertical, $\displaystyle a^2 = b^2-b^2e^2 = 36 - 16=20$.

Putting all this together, I reckon the equation is:

$\displaystyle \frac{(x-4)^2}{20} + \frac{(y+2)^2}{36} = 1$

I hope that answers some of your questions.

3. Originally Posted by ScottStedman
I was unfortunately absent on the day of graphing off-center conics in our Algebra 2 class. I've got the notes, but...I could use some help with this.
To learn, in general, how to find an ellipse's equation from given information, try some online lessons.

4. Originally Posted by Grandad
Hello ScottStedman

I hope that answers some of your questions.