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I asked the assistant proffesor on my Analytical Geometry class, but she did not know how to proof it.

Any help would be welcomed.

sorry for reposting it.

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- Thread starter sutupidmath
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- #1

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I asked the assistant proffesor on my Analytical Geometry class, but she did not know how to proof it.

Any help would be welcomed.

sorry for reposting it.

- #2

HallsofIvy

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For any ellipse, the distance, d, from one focus to the ellipse and back to the other focus is a constant.

For an ellipse with major axis a along the x-axis, minor axis b, equation

[tex]\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1[/itex]

the total distance from one focus, to the point (a, 0) to the other focus is d= 2a (that should be obvious- the distance you

The "eccentricity" is defined to be the ratio of focal distance to the length of the major-semiaxis, here that is

[tex]e= \frac{\sqrt{a^2- b^2}}{a}= \sqrt{1- \frac{b^2}{a^2}}[/tex]

The focal parameter, p, is the distance from the focus to the nearest directrix. For an ellipse, a directrix is a line perpendicular to the major axis such that the ratio of the distance from any point on to the nearest focus to the distance from that point to the nearest directrix is equal to the eccentricity. That is, with x the x coordinate of any point on the directix,

[tex]\frac{x-a}{a-c}= e[/itex]

Since c= ae, that is

[tex]\frac{x- a}{a- ae}= \frac{x- a}{a(1-e)}= e[/itex]

so that x- a= ae(1-e) and x= a+ ae(1-e)= a(1+ e- e^2).

Since x is the distance from the center of the ellipse to the directrix, p, the distance from the focus to the directrix is p= x- c= x- ae= a(1+ e- e^2)- ae= a(1- e^2). From that, a= p/(1- e

Since b

- #3

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thnx

It's all clear to me now.

It's all clear to me now.

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