Ellipse: Why is the distance from foci to y-int = a?

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masters

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This may be a simple question, but for the life of me, I can't explain why the distance from the foci to the y-int = a.

Here's the setup: Assuming the center of the ellipse is (0, 0) and the major axis lies along the x-axis and the minor axis lies along the y-axis.

The two foci are:

F[sub:3tay1x9d]1[/sub:3tay1x9d] (-c, 0) and F[sub:3tay1x9d]2[/sub:3tay1x9d] (c, 0)

The two vertices are:

V[sub:3tay1x9d]1[/sub:3tay1x9d] (-a, 0) and V[sub:3tay1x9d]2[/sub:3tay1x9d] (a, 0)

The x-intercepts are (-a, 0) and (a, 0)
The y-intercepts are (0, -b) and (0, b)

The distance from the center (0, 0) to the y-intercept is b units.
The distance from the center to a focus is c units

My question is why is the distance from the focus to the y-intercept (hypotenuse of the rt. triangle) = a?
The equation is b[sup:3tay1x9d]2[/sup:3tay1x9d] + c[sup:3tay1x9d]2[/sup:3tay1x9d] = a[sup:3tay1x9d]2[/sup:3tay1x9d] if we assume that distance is a.
 
What is the (classical) definition of an ellipse? In particular, how is an ellipse defined, in terms of the sum of the distances between the foci and any point on the ellipse?

Looking at the foci and the x-intercepts, what is the sum of the distances?

Looking at the foci and either y-intercept, noting the symmetry of this particular ellipse, what must you conclude is the distance between a y-intercept and either of the foci? :wink:

Eliz.
 
Yes, I know that the sum of the lengths of the segments drawn from each focus to a point on the ellipse is a constant. But, why is that constant 2a?
 
masters said:
Assuming the center of the ellipse is (0, 0) and the major axis lies along the x-axis and the minor axis lies along the y-axis. The two foci are: F[sub:2e3ddwk9]1[/sub:2e3ddwk9] (-c, 0) and F[sub:2e3ddwk9]2[/sub:2e3ddwk9] (c, 0). The two vertices are: [sub:2e3ddwk9]1[/sub:2e3ddwk9] (-a, 0) and V[sub:2e3ddwk9]2[/sub:2e3ddwk9] (a, 0) The x-intercepts are (-a, 0) and (a, 0) The y-intercepts are (0, -b) and (0, b)...

My question is why is the distance from the focus to the y-intercept (hypotenuse of the rt. triangle) = a?
Click to expand...
As you are ubndoubtedly aware of, an ellipse is the locus of a point, P, moving in such a way that the sum of its distances from two fixed points, F and F', called foci, is a constant.

Consider the point P moving on the end of a piece of string attached to the two foci at (-c,0) and (+c,0). Start with point P located at (-a,o). The length of that piece of string is the distance from (-c,0) to (-a,0) plus the distance from (-a,0) to (+c,0). In terms of the distances a and c, the total length of the string is (a - c) + [(a - c) + 2c] or 2a - 2c + 2c = 2a.

Since the total length of the string remains constant as point P moves from (-a,0) to (0,+b), the distance from (-c,0) to (0,+b) must be equal to "a" as is the distance from (0,+b) to (+c,0).

Therefore, b^2 + c^2 = a^2.
 
masters said:
Yes, I know that the sum of the lengths of the segments drawn from each focus to a point on the ellipse is a constant. But, why is that constant 2a?
Click to expand...
Looking at the foci and either one of the x-intercepts, what is the sum of the distances?

Since this sum is constant, and since, in this case, the foci and a y-intercept form an isosceles triangle, what must be the distance between a focus and a y-intercept?

Eliz.
 
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