Ellipse proof (1 Viewer)

[cutemouse]

Using cartesian coordinates, show that the sum of the distances from any point P on the ellipse x²/a²+y²/b² = 1 to the two foci is equal to 2a.

Thanks alot if you can help me.

 

[Trebla]

Draw a diagram of an ellipse with its foci and directrices.

Let S and S' be the positive and negative focus respectively. Let M and M' be points on the positive and negative directrix respectively such that PM and PM' are perpendicular to their corresponding directrix.

Recall the locus definition of the conic/ellipse:
PS/PM = e
where P is a point which satisfies the locus, S is a fixed point (the focus) and M is ANY point on a fixed line (the directrix)

This implies: PS = ePM
for the positive end

Similarly, for the negative end:
PS'/PM' = e
=> PS' = ePM'

Now the sum of the distances from any point P to the two foci is given by:
PS + PS' = e(PM + PM')

From your diagram it should be obvious that PM + PM' = MM' (it's just two intervals joined together), so
PS + PS' = eMM'

But MM' is the distance between the two directrices. This is equal to 2a/e (since the distance of each directrix from the y-axis is a/e)
Hence:
PS + PS' = e(2a/e)
= 2a

An alternative method, which actually uses the Cartesian coordinates, is to simply use the distance formula.

 

[cutemouse]

Hi,

Thanks for that.

To use distance formula, how would I do that?

I let P=(x,y) and S=(ae,0) and S'=(-ae,0)

So therefore PS+PS'=2a

So D_PS=sqrt((x-ae)²+y²)

I do the same thing for PS'. How would I simplify them from the square root, so that I can add them together? I know it sounds pretty novice haha, but I have no clue.

Thanks

 

[Trebla]

Hi,

Thanks for that.

To use distance formula, how would I do that?

I let P=(x,y) and S=(ae,0) and S'=(-ae,0)

So therefore PS+PS'=2a

So D_PS=sqrt((x-ae)²+y²)

I do the same thing for PS'. How would I simplify them from the square root, so that I can add them together? I know it sounds pretty novice haha, but I have no clue.

Thanks

The distance formula method is a lot longer and more complicated.

Use the fact that
b² = a²(1 - e²)
=> b²/a² = 1 - e²

So
x²/a² + y²/b² = 1
y² = b²(a² - x²)/a²
= (1 - e²)(a² - x²)
= a² - x² - a²e² + x²e²

PS = √[(x - ae)² + y²]
= √[x² - 2aex + a²e² + y²]
= √[x² - 2aex + a²e² + a² - x² - a²e² + x²e²]
= √[x²e² - 2aex + a²]
= √[(ex - a)²]
= |ex - a|
BUT the x values that satisfy the ellipse are always less than the x coordinate of the positive directrix i.e.
x < a/e
=> ex - a < 0
Hence |ex - a| = a - ex
.: PS = a - ex

Similarly:
PS' = √[(x + ae)² + y²]
= √[x² + 2aex + a²e² + y²]
= √[x² + 2aex + a²e² + a² - x² - a²e² + x²e²]
= √[x²e² + 2aex + a²]
= √[(ex + a)²]
= |ex + a|
BUT since the x values that satisfy the ellipse are always greater than the x coordinate of the negative directrix i.e.
x > - a/e
=> ex + a > 0
Hence |ex + a| = ex + a
.: PS' = ex + a

So
PS + PS' = a - ex + ex + a
= 2a