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Conic section problem (ellipse) 
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#1
Nov2012, 09:56 AM

P: 264

1. The problem statement, all variables and given/known data
Let [itex] k>0 [/itex] be such that [itex] (x^2x)+k(y^2y)=0 [/itex] defines an ellipse with focal length equal to [itex] 2 [/itex]. If [itex] (p,q) [/itex] are the coordinates of a point in the ellipse with [itex] q^2  q\not=0 [/itex], then what is [itex] \frac{pp^2}{q^2q} [/itex]? 2. Relevant equations The fact that the sum of the distances from any point in an ellipse is equal to two times the length of the major axis. 3. The attempt at a solution First, I wrote the ellipse equation in "standard" form: [tex] \frac{(x\frac{1}{2})^2}{\frac{1+k}{4}}+\frac{(y\frac{1}{2})^2}{\frac{1+k}{4k}}=1 [/tex] Which means that the ellipse is centered at [itex] (\frac{1}{2},\frac{1}{2}) [/itex] and it's major axis is the xaxis. Because [itex] k [/itex] could never be less than [itex] 1 [/itex] since that would make the major axis be less than the specified focal length. And as far as I know, that is not allowed. Anyway, I could solve for [itex] \frac{1+k}{4} [/itex] and [itex] \frac{1+k}{4k} [/itex] by using the fact that major axis of the ellipse squared is equal to the minor axis squared plus half the focal length squared, which gives: [tex] (\frac{1+k}{4})^2= (\frac{1+k}{4k})^2+1 [/tex] Then I could say that the sum of the distances from any point to each focus of the ellipse is always equal to [itex] 2(\frac{1+k}{4}) [/itex]. And that leaves the equation below to solve for [itex] \frac{pp^2}{q^2q} [/itex]: [tex] \sqrt{(\frac{3}{2}+p)^2+q^2} + \sqrt{(\frac{5}{2}+p)^2+q^2}=2(\frac{1+k}{4}) [/tex] Where [itex] k [/itex] is a known. Okay, now I am definitely stuck.. So I have a few questions. First; can I find the relationship of p and q in the expression [itex] \frac{pp^2}{q^2q} [/itex] just by knowing their relationship in [itex] \sqrt{(\frac{3}{2}+p)^2+q^2} + \sqrt{(\frac{5}{2}+p)^2+q^2} [/itex]? Second; why is it that [itex] \frac{pp^2}{q^2q} [/itex] is a constant ratio for every point in the ellipse? I feel like that is some property of ellipses which if I knew it would make the problem way easier. Is that right? Third; if I can't just do what I said above would I actually have to do something else? What? Thanks! 


#2
Nov2012, 11:28 AM

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P: 12,113

How do you get those expression in the square root?
With [itex] (x^2x)+k(y^2y)=0 [/itex] and replacing x>p, y>q for points on the boundary, we know that ##(p^2p)+k(q^2q)=0## which gives [itex] \frac{pp^2}{q^2q} =k[/itex] Seen as function of x, [itex] (x^2x)+k(y^2y) [/itex] is a parabola with two roots (corresponding to points on the ellipse) and negative values in between. Therefore, each point inside has [itex] (x^2x)+k(y^2y) < 0[/itex] with a minimum of [itex] \frac{1}{4}+k(y^2y) [/itex] at x=1/2 (the center of the ellipse). Similar arguments should lead to an upper bound on [itex] \frac{pp^2}{q^2q}[/itex]. 


#3
Nov2012, 12:03 PM

P: 264

But that is useless now.. I think. I tried using the method I mentioned, now assuming that the answer I am looking for is k, but I can't get to any results that resemble the ones I can choose from. Is there something wrong with my approach to finding [itex] k [/itex]? EDIT: AH! I got it now! The thing that I was using wrong was that in the equation [tex] \frac{(x\frac{1}{2})^2}{\frac{1+k}{4}}+\frac{(y\frac{1}{2})^2}{\frac{1+k}{4k}}=1 [/tex] The major and minor axis were actually the square roots of the denominators there. 


#4
Nov2012, 05:45 PM

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P: 12,113

Conic section problem (ellipse)
I got confused by "point in the ellipse". If you mean "point on the ellipse", ignore the second part.
And you are right, the sign in my post is wrong. The fraction is always k for all points of the ellipse, which follows directly from its equation. 


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