# Conics

Discussion in 'General Off-Topic Chat' started by Quantum, May 14, 2007.

# Conics

937 Views
1. ### QuantumGBAtemp Regular

Member
2
Jun 12, 2004
Ohio
So I have this math problem:

Using the ellipse x^2/a^2 + y^2/b^2 = 1 and assuming that a>b, determine two formulae that calculate the distance from a point on the ellipse to each of the foci. Use the formulae to show that the sum of the distances is 2a.

I know that the foci are at (sqrt(a^2-b^2), 0) and (-sqrt(a^2-b^2), 0), but I can't get a formula for the distance of a point to them. I was able to get this extremely long and ugly value, but I'm unable to simplify it and show that it will add to 2a. Any suggestions? Thanks in advance.

2. ### FAST6191Techromancer

Reporter
21
Nov 21, 2005
I am in a bit of a rush right now so I may have to post halfway through and run.

Elliptical maths huh, given this is some of the hardest maths there is (it makes for some hellish encryption) I do not know how much help I can be (luckily you are only a few steps up from circle stuff and still fairly simplified) but here goes:

You have foci, call these are you origins. You now want a point on the ellipse:
Simple rearrangement of the equation you have for this ellipse:
y value: y^2=b^2(1-x^2/a^2)

x value: x^2=a^2(1-y^2/b^2) (you likely will not need this but hey)

leave them squared for reasons that will become obvious very soon.

It may help to have a graph here if you are not so hot visualising stuff, between the foci and the point you have a line, it is also a right angled triangle therefore Pythagoras appears.

pythag equation:
hyp^2=opp^2+adj^2
Hyp is obviously the length of the line from from the foci to the point, it really does not matter what you call the others as long as one does for y and the other for x.

Now it is my understanding that a and b are related to the values for opp and adj.
b - (b-y) (naturally if the loci were not at y-0 then it gets a bit more fun).

you need a value for the "x direction" now, your foci do not sit right here for me (I am thinking it should be eccentricity x a with eccentricity being root 1 minus b squared over a squared) unless of course they were defined like that in which case it is all good.

Still assuming you have a foci point you need the length between the x value of the point on the ellipse and said foci (be very careful with sign when you do it for the opposite foci, quite in fact I suggest taking absolute values for the first run through).

It would seem I have ran out of time so I can not go through it all but you should now have all you need for distance from a point to foci, summing the equations and paying special attention to signs should allow you to cancel stuff (setting b to 1 should help immensely).

3. ### OSWWii King

Former Staff
2
Oct 30, 2006
Maybe we should make a general maths topic... since i have problems too sometimes.
but sorry since i suck i can't help you

4. ### PsyfiraCredit: 0ml. Insert tea to continue

Member
3
Dec 31, 2003
England
*Screams, throws hand over her head and runs away very very fast with flashbacks of 3rd year software project ellipse hell* I spent a couple of months getting held up just trying to get the right f'ing ellipse formulas for what I was doing, never did get it working in the end (though I came close). Despite all that I didn't need anything to do with the foci so I can't help you there