An ellipse fitting into a notch problem.

[polishdude20]

Homework Statement

I did this one before a few weeks ago but now I can't seem to get the right answer:

An elliptical disk is to fit snuggly and squarely into a notch cut into a rectangular plate. The notch is 180 mm wide. If the disk's major axis is 280 mm long and is parallel to the long edge of the notch, and the disk's minor axis is 40 mm wide. How deep does the notch need to be for the right edge of the disk to just touch the back of the notch?

Homework Equations

(x^2)/(a^2) + (y^2)/(b^2) = 1

The Attempt at a Solution

So I know that the "A" value is half of the major axis of the ellipse so it's 140mm, the "B" value is half of the minor axis so it's 20mm. The notch is 180mm wide so the x value for the edge of the notch which touches the ellipse is half of that which is 90mm. So I lug the numbers into the ellipse equation to solve for the Y value. I square the 140 and the 20 and the 90 in my calculations according to the formula and I get -12.8 so I thought that that would be the distance from the centre not from the bottom so I subtracted 12.8 from 20 which gave me 7.1. The answer is wrong.

Anyone have any ideas?

 

[NascentOxygen]

I square the 140 and the 20 and the 90 in my calculations according to the formula and I get -12.8

You get -12.8 before you take the square root to give y? Then there is probably something amiss right there. You explained what you were doing, and it seems right, so you'd better recheck. Post the arithmetic expression if you can't see your mistake.

so I thought that that would be the distance from the centre not from the bottom

Take a look at the diagram you drew, it will show exactly what y is. (You did draw a neat, large diagram, didn't you? With the ellipse centred at (0,0)? )

What is the correct answer, were you told?