Extensions Page 2b.2

Page 2b.2

Deriving the Equation for Focal Length: c = (b² - a²)^½

Recall from page 2b.1 that when the horizontal axis was longer than the verticle axis, the sum of the distances between any point on the ellipse, P, and each of the foci was equal to the length of the major axis, 2a.

In the case we are now considering, the vertical axis fo the ellipse is longer than the horizontal axis. It can be shown that in this case, the sum of the distances from point P to each focal point is 2b, the length of the vertical axis.

Now let P be the left endpoint of the horizontal axis, the point A₂. PF₁ + PF₂ = A₂F₁ + A₂F₂

By symmetry, the triangle A₂F₁F₂ is an isocoles triangle. Therefore, A₂F₁ = A₂F₂ Thus: A₂F₁ + A₂F₂ = 2 A₂F₁

As shown above, this sum must equal 2b, so : 2 A₂F₁ = 2b

Divide both sides by 2: A₂F₁ = b

So now we know that the distance from the end of the vertical axis to either focal point is b.

Figure 1

Now apply the Pythagorean Theorem to the triangle OA₂F₁:
a = OA₂; b = A₂F₁; c = OF₁ b² = a² + c²

Solve this equation for c, taking only the postive root: c² = b² - a²
==> c = (b² - a²)^½
We have once again found an expression for c, the focal length. c = (b² - a²)^½

Recall from page 2b.1 that when the horizontal axis was longer than the verticle axis, the sum of the distances between any point on the ellipse, P, and each of the foci was equal to the length of the major axis, 2a.
In the case we are now considering, the vertical axis fo the ellipse is longer than the horizontal axis. It can be shown that in this case, the sum of the distances from point P to each focal point is 2b, the length of the vertical axis.
Now let P be the left endpoint of the horizontal axis, the point A₂.	PF₁ + PF₂ = A₂F₁ + A₂F₂
By symmetry, the triangle A₂F₁F₂ is an isocoles triangle. Therefore, A₂F₁ = A₂F₂ Thus:	A₂F₁ + A₂F₂ = 2 A₂F₁
As shown above, this sum must equal 2b, so :	2 A₂F₁ = 2b
Divide both sides by 2:	A₂F₁ = b
So now we know that the distance from the end of the vertical axis to either focal point is b.

Now apply the Pythagorean Theorem to the triangle OA₂F₁: a = OA₂; b = A₂F₁; c = OF₁	b² = a² + c²
Solve this equation for c, taking only the postive root:	c² = b² - a² ==> c = (b² - a²)^½
We have once again found an expression for c, the focal length. c = (b² - a²)^½