Ellipse: sum of distances from foci to point on ellipse is 2a

Nazariy

Junior Member
Joined
Jan 21, 2014
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124
Hello,

Basically I was trying to prove the equation of ellipse by equating the sum of distances from foci to point on ellipse and sum of distances from foci to far left point of ellipse (i.e. > (-a,0)). But that does not seem to work perfectly.

There is an easier way to prove the ellipse equation, which is equate the sum of distances from foci to point on ellipse with 2a. The question I have is where does that 2a come from??

Thanks
 
Hello,

Basically I was trying to prove the equation of ellipse by equating the sum of distances from foci to point on ellipse and sum of distances from foci to far left point of ellipse (i.e. > (-a,0)). But that does not seem to work perfectly.

There is an easier way to prove the ellipse equation, which is equate the sum of distances from foci to point on ellipse with 2a. The question I have is where does that 2a come from??

Thanks

What is the distance between two focii of an ellipse?

Look up definition of "ellipse" w.r.t. focii.

Then come beack and tell us what you found.
 
What is the distance between two focii of an ellipse?

Look up definition of "ellipse" w.r.t. focii.

Then come beack and tell us what you found.

I do not want to sound rude, but have I not known the definition of an ellipse would I be able to actually prove the equation?

It is following: the sum of distances from foci to any point lying on the ellipse is constant. Does this have to help me with thinking why it is 2a?
 
I do not want to sound rude, but have I not known the definition of an ellipse would I be able to actually prove the equation?

It is following: the sum of distances from foci to any point lying on the ellipse is constant. Does this have to help me with thinking why it is 2a?

You have claimed you have proved "something" - but you have not shown any work!

Anyway - you need another parameter (like minor axis) - to get the equation you are looking for.
 
Hello,

Basically I was trying to prove the equation of ellipse by equating the sum of distances from foci to point on ellipse and sum of distances from foci to far left point of ellipse (i.e. > (-a,0)). But that does not seem to work perfectly.

There is an easier way to prove the ellipse equation, which is equate the sum of distances from foci to point on ellipse with 2a. The question I have is where does that 2a come from??
If the major axis is horizontal and the center is \(\displaystyle (0,0)\) then the two major vertices are \(\displaystyle (a,0)~\&~(-a,0)\) where \(\displaystyle a>0\).
The two minor vertices are \(\displaystyle (0,b)~\&~(0,-b)\) where \(\displaystyle a>b>0\).

The two focii are \(\displaystyle F_1: (c,0)~\&~F_2: (-c,0)\) where \(\displaystyle a>c>0\).

If the point \(\displaystyle P: (p,q)\). is on the ellipse then \(\displaystyle F_1P+F_2P=K\) or
\(\displaystyle \sqrt{(p-c)^2+q^2}+\sqrt{(p+c)^2+q^2}=K\) show that \(\displaystyle K=2a\) by letting \(\displaystyle p=a~\&~q=0~.\)
 
You have claimed you have proved "something" - but you have not shown any work!

Anyway - you need another parameter (like minor axis) - to get the equation you are looking for.

Correction: "attempt to prove", I thought it was obvious. I mean it is clear from my initial statement that I do understand what the definition of ellipse is. Sorry, I do not want to sound rude.
 
If the major axis is horizontal and the center is \(\displaystyle (0,0)\) then the two major vertices are \(\displaystyle (a,0)~\&~(-a,0)\) where \(\displaystyle a>0\).
The two minor vertices are \(\displaystyle (0,b)~\&~(0,-b)\) where \(\displaystyle a>b>0\).

The two focii are \(\displaystyle F_1: (c,0)~\&~F_2: (-c,0)\) where \(\displaystyle a>c>0\).

If the point \(\displaystyle P: (p,q)\). is on the ellipse then \(\displaystyle F_1P+F_2P=K\) or
\(\displaystyle \sqrt{(p-c)^2+q^2}+\sqrt{(p+c)^2+q^2}=K\) show that \(\displaystyle K=2a\) by letting \(\displaystyle p=a~\&~q=0~.\)

Thank you pka! You are extremely helpful!
 
If the major axis is horizontal and the center is \(\displaystyle (0,0)\) then the two major vertices are \(\displaystyle (a,0)~\&~(-a,0)\) where \(\displaystyle a>0\).
The two minor vertices are \(\displaystyle (0,b)~\&~(0,-b)\) where \(\displaystyle a>b>0\).

The two focii are \(\displaystyle F_1: (c,0)~\&~F_2: (-c,0)\) where \(\displaystyle a>c>0\).

If the point \(\displaystyle P: (p,q)\). is on the ellipse then \(\displaystyle F_1P+F_2P=K\) or
\(\displaystyle \sqrt{(p-c)^2+q^2}+\sqrt{(p+c)^2+q^2}=K\) show that \(\displaystyle K=2a\) by letting \(\displaystyle p=a~\&~q=0~.\)

No idea what to do next.. Feeling silly

Capture.GIF
 
I had hoped that you would have noted that the length of the major axis is \(\displaystyle 2a\).
The sum of distances from the focii to \(\displaystyle (a,0)\) has to be \(\displaystyle 2a\).

Yes, indeed, I do realize it conceptually.

No matter whether we derive the length from focii to (-a,0) or to (a,0), it is evident that the length is 2a, because we have a length from further foci + length from foci that is closer = the whole x range that is captured by ellipse, which is absolute value of -a-a, thus 2a.

I would however like to see that mathematically. Also, if I have problems with maths, I have to take the problem to full completion and to understand it fully. So I will still try to figure out how to solve that equation.

There was a file I was looking at yesterday, where someone derived the equation of an ellipse and they used some substitution for a^2 + c^2 (or was it some other combination of a^2,b^2,c^2) from Pythagorean theorem which I thought was a bit odd, but perhaps may help me if I consider it further. So I will now consider this option, perhaps it will help me tackle the problem further. I have to solve it mathematically.
 
I had hoped that you would have noted that the length of the major axis is \(\displaystyle 2a\).
The sum of distances from the focii to \(\displaystyle (a,0)\) has to be \(\displaystyle 2a\).

Ok probably silly to try and figure out that that length is 2a. Can't seem to be able to do that... But I am sure it is super simple and I am missing something...
 
Ok probably silly to try and figure out that that length is 2a. Can't seem to be able to do that.

We know that \(\displaystyle a > 0\) and we know that \(\displaystyle (a,0)~\&~(-a,0)\) are the endpoints of the major axis.
Therefore, the length of the major axis is \(\displaystyle 2a\).
 
We know that \(\displaystyle a > 0\) and we know that \(\displaystyle (a,0)~\&~(-a,0)\) are the endpoints of the major axis.
Therefore, the length of the major axis is \(\displaystyle 2a\).

Oh Lord... Sometimes I just overheat... haha, sorry.

Hmm, that's what I get when I equate sum of lengths from focii to P(x,y) that lies on ellipse with 2a.
aa.GIF which is,
bb.GIF
Which makes me think I made a mistake somewhere with a sign and am also missing 1 above.
 
We know that \(\displaystyle a > 0\) and we know that \(\displaystyle (a,0)~\&~(-a,0)\) are the endpoints of the major axis.
Therefore, the length of the major axis is \(\displaystyle 2a\).

Found three mistakes in those guy's derivations which cancel out, and make him derive the correct formula..
 
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