# Conics Problem Part 3

## Homework Statement

http://img103.imageshack.us/img103/3784/arch3tf0.jpg [Broken]​
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On picture above

## The Attempt at a Solution

Again, I just want someone to check my work.

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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.

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.

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

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.

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

Do you mean the center?

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).