# Math Help - Ellipse

1. ## Ellipse

A hallway 5.0 m wide has a ceiling whose cross section is a semi ellipse. The ceiling is 3.0 m high at the walls and 4.0 m high at the center. Find the height of the ceiling 1.0 m from each wall.

How would I start this problem?

2. Originally Posted by OzzMan
A hallway 5.0 m wide has a ceiling whose cross section is a semi ellipse. The ceiling is 3.0 m high at the walls and 4.0 m high at the center. Find the height of the ceiling 1.0 m from each wall.

How would I start this problem?
From the measures of the hallway you know about the ellipse:

semi-major axis : 2.5 m
semi-minor axis : 1 m
coordinates of the center: C(0, 3)

Thus the equation of the ellipse is:

$\frac{x^2}{\left(\frac52\right)^2}+\frac{(y-3)^2}{1^2} = 1$

Because you only need the upper part of the ellipse you can solve for y to get the equation of a function:

$e(x) = 3+\sqrt{1-\frac{4x^2}{25}}$

The points A and B are 1 m from the wall: A(-1.5, e(1.5)) and B(1.5, e(1.5))
Because the ellipse is symmetric about the y-axis it is only necessary to calculate e(1.5).

I've got A(-1.5, 3.8), B(1.5, 3.8)

3. Oh I see. So my answers would be in the form of (x,y). Right?

4. Originally Posted by OzzMan
Oh I see. So my answers would be in the form of (x,y). Right?
Not necessarily. You rae asked to calculate the height. It would be sufficient to give the answer as h = 3.8 m

5. Originally Posted by earboth
$e(x) = 3+\sqrt{1-\frac{4x^2}{25}}$
Shouldn't it be $e(y) = 3+\sqrt{1-\frac{4x^2}{25}}$

And also how come you have to put an e in there?

6. Originally Posted by OzzMan
Shouldn't it be $e(y) = 3+\sqrt{1-\frac{4x^2}{25}}$

And also how come you have to put an e in there?
I transformed the equation of the ellipse into the equation of a function whose graph is the upper part of the ellipse.

Therefore:

$y = e(x) = 3+\sqrt{1-\frac{4x^2}{25}}$

which will yield the height above the x-axis.