1. ## Ellipses

I've been trying to find a way into solving this problem for about an hour, but nothing's clicking. The only information given is its length along the major axis and a point on the ellipse.
Here it is:

A bridge is to be built in the shape of a semielliptical arch and is to have a span of 100 feet. The height of the arch, at a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.

So I know that A squared is 2500, but I can't figure out how to use the point (40,10) to arrive at the focus, which would then let me figure out the height.

2. Originally Posted by spiritualfields
I've been trying to find a way into solving this problem for about an hour, but nothing's clicking. The only information given is its length along the major axis and a point on the ellipse.
Here it is:

A bridge is to be built in the shape of a semielliptical arch and is to have a span of 100 feet. The height of the arch, at a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.

So I know that A squared is 2500, but I can't figure out how to use the point (40,10) to arrive at the focus, which would then let me figure out the height.
I dont know what a semielliptical arch is...but I assume your answer has to do with simiar triangles...

some else might be more help...I got up at 3:15 this morn to do homework and i'm wiped...

dan

3. Originally Posted by spiritualfields
I've been trying to find a way into solving this problem for about an hour, but nothing's clicking. The only information given is its length along the major axis and a point on the ellipse.
Here it is:

A bridge is to be built in the shape of a semielliptical arch and is to have a span of 100 feet. The height of the arch, at a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.

So I know that A squared is 2500, but I can't figure out how to use the point (40,10) to arrive at the focus, which would then let me figure out the height.
The equation for an ellipse is $x^2/a^2 + y^2/b^2 = 1.$ You know $a = 100/2.$ Plug that and $(x,y) = (40,10)$ into the equation and solve for $b.$

4. Originally Posted by spiritualfields
I've been trying to find a way into solving this problem for about an hour, but nothing's clicking. The only information given is its length along the major axis and a point on the ellipse.
Here it is:

A bridge is to be built in the shape of a semielliptical arch and is to have a span of 100 feet. The height of the arch, at a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.

So I know that A squared is 2500, but I can't figure out how to use the point (40,10) to arrive at the focus, which would then let me figure out the height.
The ellipse you speak of can be thought to be centered on (0, 0) with either end of the bridge at x = $\pm$50 ft. The equation of such an ellipse, with the central long axis on the x-axis is:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

You know the value of the semi-axis major, a = 50. You also know a point on the ellipse: (40, 10) (You actually know the point (-40, 10) as well, since the ellipse is symmetrical.)

Plug all this into your equation:

$\frac{40^2}{50^2} + \frac{10^2}{b^2} = 1$
This is an equation for your semi-axis minor, b, which tells you the height of the ellipse (bridge) above the origin (center of the bridge.)

-Dan

5. Boy, do I feel dumb. Thanks for the explanation. It seems so simple now.

6. Originally Posted by spiritualfields
Boy, do I feel dumb. Thanks for the explanation. It seems so simple now.
It almost always seems simple after you have the answer... Don't be too hard on yourself.

-Dan