a friend told me about this website and said people were very helpful. so I'm gonna give it a shot. I'm having some trouble with a few trig word problems.

A reflecting telescope has a mirror shaped paraboloid of revolution. If the distance from the vertex to the focus is 26 feet and the distance across the top of the mirror is 58 inches, how deep is the mirror in the center?

I can't get a grasp on how I should draw this to set it up.. or if I even need to..

2. Originally Posted by NKS
a friend told me about this website and said people were very helpful. so I'm gonna give it a shot. I'm having some trouble with a few trig word problems.

A reflecting telescope has a mirror shaped paraboloid of revolution. If the distance from the vertex to the focus is 26 feet and the distance across the top of the mirror is 58 inches, how deep is the mirror in the center?

I can't get a grasp on how I should draw this to set it up.. or if I even need to..
The cross section through the vertex is a parabola. Define a coordinate system with the vertex at V(0, 0) and the focus at F(0, 26).

Then the parabola has the equation:

$4 \cdot 26 \cdot y = x^2$

The rim of the mirror is 29 inches away from the axis of the parabola:

$4 \cdot 26 \cdot y = 29^2$

Solve for y which will yield the depth of the mirror.

3. HA it seems so easy now that I look back on it.. Thanks a bunch. I think my problem has been that I keep associating paraboliods with the equation to an ellipse. I keep forgetting that they have their own equation. I got to remember that a semi-ellipse and a parabola are two different things. thanks again!!

4. hmm I noticed something as I was looking over the set up to help with other problems..

wouldn't you need to convert the 26ft to inches. it would be 312 * 4 * y = 29^2
correct?

5. Originally Posted by NKS
hmm I noticed something as I was looking over the set up to help with other problems..

wouldn't you need to convert the 26ft to inches. it would be 312 * 4 * y = 29^2
correct?
Of course, you are right.

For me all calculations with inches, feet, yards, fathoms, miles, etc. are very exotic exercises because I'm used only to the very simple system of centimeter, meter and kilometer.

But the way to solve the question doesn't change. I now have a depth of the mirror of 0.674''.