Locus of points making an ellipse

[fireflies]

I know that

1) when eccentricity is less than 1 then it is an ellipse

2) locus of points making sum of the distance from two fixed points(foci) with that point a constant, creates ellipse.

Here comes the question, I understand that locus made according to number 2, is ellipsoidal. But how can it give the same equation of an ellipse? Or in reverse way how the sum of the distance of any point on the ellipse from the foci is constant?

 

[jedishrfu]

You might be able to derive the equation for an ellipse for a specific example.

Start with foci at (-3,0) and (3,0) and then use the pythagorean theorem to get the distance from point on the ellipse and the two foci then add them together. Next do it again using a point (x,y).

 

[fireflies]

I tried it. I considered a normal ellipse with an equation

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

taking eccentricity e, foci comes (ae,0) and (-ae,0). On, (0,b) applying Pythagorus' theorem I got

(ae)^2 + b^2 = s^2

again at vertex (a,0) there is

2S=2a+ae
or, S= a+ ae/2
or, S^2=a^2 +a^2e + (ae/2)^2

Well, then I did quite different thing. I put the two values of s^2 together to get that if the equation is correct.

That brought

a^2 + a^2e + (ae/2)^2 = a^2 (e^2 + (b/a)^2)

or, 1+ e + e^2/4 = 1 - (b/a)^2 + (b/a)^2

or, e + e/4= O

But it cannot be. There must be some problem in calculation. And I don't know how else to solve it

 

[mathman]

I tried it. I considered a normal ellipse with an equation

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

taking eccentricity e, foci comes (ae,0) and (-ae,0). On, (0,b) applying Pythagorus' theorem I got

(ae)^2 + b^2 = s^2

again at vertex (a,0) there is

2S=2a+ae {I get 2s=(a+ae)+(a-ae)=2s}
or, S= a+ ae/2
or, S^2=a^2 +a^2e + (ae/2)^2

Well, then I did quite different thing. I put the two values of s^2 together to get that if the equation is correct.

That brought

a^2 + a^2e + (ae/2)^2 = a^2 (e^2 + (b/a)^2)

or, 1+ e + e^2/4 = 1 - (b/a)^2 + (b/a)^2 {where did -(b/a)^2 come from?}

or, e + e/4= O

But it cannot be. There must be some problem in calculation. And I don't know how else to solve it

My comments in text using {}.

 

[fireflies]

e^2= 1-(b/a)^2

when a>b

 

[fireflies]

Oh yes, s=a then, 2s = 2a.

so, a^2=a^2(e^2+(b/a)^2)

which brings 1=1.

So, the calculation is true.

I got the solution.

But how can we possibly come to conclusion that any other point P(x,y) also give the same 2S=2a?

Should I try finding out for any such point, or, is there an easy conclusion from the upper two cases?

How did anyone who first found it out do it then?

 

[fireflies]

Here 2S means S+S' (one from each focus)

 

[jedishrfu]

Here's an interesting discussion on ellipses:

https://en.wikipedia.org/wiki/Ellipse

 

[fireflies]

Here's an interesting discussion on ellipses:

https://en.wikipedia.org/wiki/Ellipse

Yes, I've found the whole proof from the page giving another link.

(The link is: https://en.m.wikipedia.org/wiki/Proofs_involving_the_ellipse )Thanks!