# Conics Problem Part 3

1. Jun 12, 2008

### temaire

1. The problem statement, all variables and given/known data

2. Relevant equations
On picture above

3. The attempt at a solution
Again, I just want someone to check my work.

2. Jun 12, 2008

### konthelion

For the first problem on this page, (again from the previous problem) for a hyperbola, $$c^2=a^2+b^2$$

c=distance from center to focii, a = distance from center to vertex
Incidentally, you don't need to calculate for a since it's given.

3. Jun 13, 2008

### konthelion

Whoops, I realized that the problem is using a different form of the vertical hyperbola instead of $$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$$, so in this case switch a and b. In the other problem, where b=10 is correct.

Last edited: Jun 13, 2008
4. Jun 13, 2008

### temaire

For question 2, do you think I was right to mention the (h,k) values as a difference between the two graphs?

5. Jun 13, 2008

### konthelion

Since (h,k) in the parabolic model is the vertex whereas the (h,k) in the semi-elliptical model is the center, I believe you should also compare the focus of the parabola and the focii of the ellipse.

6. Jun 13, 2008

### temaire

We haven't used the terms focus or focii with these problems in class. What do you mean by them?

7. Jun 13, 2008

### temaire

Do you mean the center?

8. Jun 13, 2008

### konthelion

The focii for an ellipse is the point that lies on the major axis(the longer side/axis) of the ellipse. There are two focii in this ellipse. (h+c,k) and (h-c,k). In an ellipse(for both vertial and horizontal ellipses), $$b^2=a^2-c^2$$, where a is always the large axis and b is the smaller axis.

In a parabola, the there is only one focus. Since this parabola opens down, then the focus is at (h,-c+k).