Definition 1 implies Definition 3.
This has a beautiful demonstration invented in 1822 by the Belgian mathematician G. P. Dandelin. The two points are known as the foci of the ellipse, and the constant sum is equal to the longest diameter of the ellipse - its major axis. Definition 3 is also known as the focal property of the ellipse.
Definition 3 implies Definition 2.
Let the x-axis run through the foci, with the origin midway between the two. The foci then will have coordinates E(-c, 0) and F(c, 0). Let P(x, y) be a generic point on the ellipse, so that |EP| + |FP| = 2a = const and a > c. The segments EP and FP are known as the focal radii of P, the left and the right radii, correspondingly. So we start with
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√(x + c)² + y² + √(x - c)² + y² = 2a.
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Move one of the square roots to the right and square both sides:
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(x + c)² + y² = (x - c)² + y² - 4a√(x - c)² + y² + 4a²
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which simplifies to
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a² - xc = a√(x - c)² + y².
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With an additional squaring eliminate the radical:
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a4 - 2a2xc + x2c2 = a2(x2 - 2xc + c2 + y2).
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A simplification yields
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a²(a² - c²) = (a² - c²) x² + a²y².
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Introducing b² = a² - c² we get equation (1).
a, b, c are called the major semiaxis, the minor semiaxis, and the linear eccentricity, respectively. 2c is the focal distance.
Points (±a, 0) and (0, ±b) lie on the ellipse and are known as its vertices. e = c/a is called the eccentricity. For ellipse, 0 ≤ e < 1.
Definition 2 implies Definition 4.
We may now find |EP| and |FP| separately. Say,
| | |EP|² | = (x + c)² + y² |
| | | = (x + c)² + b²(1 - x² / a²) |
| | | = (1 - b² / a²) x² + 2xc + c² + b² |
| | | = x² c² / a² + 2xc + a² |
| | | = e² x² + 2xea + a² |
| | | = (a + ex)². |
Similarly, |FP|² = (a - ex)². Since |x| ≤ a and hence |xe| < a,
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|EP| = a + ex,
|FP| = a - ex.
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Now, consider the straight line d: x = -a/e perpendicular to the major axis. The distance between P(x, y) and that line is
making the ratio of the distances from P to E and to d equal e < 1. The line d is called the (left) directrix of the ellipse. There is the right directrix, of course, whose equation is x = a/e. It plays the role similar to the left one but with respect to the right focus F.
Definition 4 implies Definition 2.
Let e be the ratio of distances and choose the system of coordinates so that the focus and the related directrix be E(-ae, 0) and x = -a/e. Let P(x, y) be a generic point on the locus in question. Then, squaring the ratio of the distances to the point and to the line gives
| (2) |
[(x + ae)² + y²] / (x + a/e)² = e².
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Multiply out and simplify to obtain
| (3) |
x²(1 - e²) + y² = a²(1 - e²)
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which, upon setting b² = a² (1 - e²), leads to the equation (1) in Definition 2.
Definition 4 implies Definition 3.
We may change the sign of x in equation (2) to obtain
| (2') |
[(x - ae)² + y²] / (x - a/e)² = e².
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which tells us that in Definition 4 there is a second pair of point-line with equations F(ae, 0) and x = a/e. (Naturally (2') leads to the same equation (3).)
From (2) and (2'),
| (4) |
| |EP| + |FP| | = |e(x - a/e)| + |e(x + a/e)| |
| | = |a - xe| + |a + xe|. |
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Now observe, that for e to be less than 1, the whole locus has to be to the left of the directrix x = a/e (and to the right of x = -a/e.) For, otherwise, P would be father away from F than from the line x = a/e. This remark allows us to drop the symbols of absolute value in (4):
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| |EP| + |FP| | = |a - xe| + |a + xe| |
| | = a - xe + a + xe |
| | = 2a. |
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Definition 2 implies Definition 5.
Assuming a > b, let L be the x-axis and k = b/a.
Definition 5 implies Definition 2.
Assuming 1 > k > 0, take L as the x-axis and let b = kr, where r is the radius of circle C. The locus of point P' will then have equation (1). If k is negative, the curve also reflects around L. For k > 1, the circle is rather stretched than squashed; it is squashed in the direction perpendicular to L.
... to be continued ...
