Extensions Page 2b.1

Page 2b.1

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

According to the Locus definition of an ellipse, the sum of the distances from any point, P, on the ellipse to each of the focal points, F₁ and F₂, is constant: PF₁ + PF₂ = constant where PF₁ is the notation for the length of the line segment connecting points P and F₁.

The point P could be at any point on the ellipse, but let's consider the case when P equals A₁, where A₁ is the rightmost endpoint of the horizontal axis. PF₁ + PF₂ = A₁F₁ + A₁F₂
Let A₂ be the leftmost endpoint of the horizontal axis. Then, by symmetry of the ellipse, A₁F₁ = A₂F₂.
Thus, we can rewrite the statement as: A₁F₁ + A₁F₂ = A₂F₂ + A₁F₂
But A₂F₂ + A₁F₂ is actually the entire horizontal axis. Therefore: A₂F₂ + A₁F₂ = A₁A₂ = 2a

By definition, we knew that the sum of the distances from any point P to each of the focal points was a constant, but now we know even more. The sum of these distance always equals 2a, the length of the horizontal axis.

Now, by cleverly choosing P as another special point on the ellipse we can get closer to showing our aim, finding an equation for the locations of the focal points.
Let P be located at the endpoint of the vertical axis, at the point B₁. (See Figure 1(b)) PF₁ + PF₂ = B₁F₁ + B₁F₂

By symmetry, the triangle B₁F₁F₂ is an isocoles triangle. Therefore, B₁F₁ = B₁F₂ Thus: B₁F₁ + B₁F₂ = 2 B₁F₁

As shown above, this sum must equal 2a, so: 2 B₁F₁ = 2a

Divide both sides by 2: B₁F₁ = a = B₁F₂

We now know that the length from the end of the vertical axis to either focal point is a.

Figure 1

Now apply the Pythagorean Theorem to the triangle OB₁F₁, where O is the centre of the ellispe:
a = B₁F₁; b = OB₁; c = OF₁ a² = b² + c²
Solve this equation for c, taking only the positive root: c² = a² - b²
==> c = (a² - b²)^½
Finally, we have found an expression for c, the focal length. c = (a² - b²)^½

According to the Locus definition of an ellipse, the sum of the distances from any point, P, on the ellipse to each of the focal points, F₁ and F₂, is constant: PF₁ + PF₂ = constant where PF₁ is the notation for the length of the line segment connecting points P and F₁.
The point P could be at any point on the ellipse, but let's consider the case when P equals A₁, where A₁ is the rightmost endpoint of the horizontal axis.	PF₁ + PF₂ = A₁F₁ + A₁F₂
Let A₂ be the leftmost endpoint of the horizontal axis. Then, by symmetry of the ellipse, A₁F₁ = A₂F₂. Thus, we can rewrite the statement as:	A₁F₁ + A₁F₂ = A₂F₂ + A₁F₂
But A₂F₂ + A₁F₂ is actually the entire horizontal axis. Therefore:	A₂F₂ + A₁F₂ = A₁A₂ = 2a
By definition, we knew that the sum of the distances from any point P to each of the focal points was a constant, but now we know even more. The sum of these distance always equals 2a, the length of the horizontal axis.

Now, by cleverly choosing P as another special point on the ellipse we can get closer to showing our aim, finding an equation for the locations of the focal points.
Let P be located at the endpoint of the vertical axis, at the point B₁. (See Figure 1(b))	PF₁ + PF₂ = B₁F₁ + B₁F₂
By symmetry, the triangle B₁F₁F₂ is an isocoles triangle. Therefore, B₁F₁ = B₁F₂ Thus:	B₁F₁ + B₁F₂ = 2 B₁F₁
As shown above, this sum must equal 2a, so:	2 B₁F₁ = 2a
Divide both sides by 2:	B₁F₁ = a = B₁F₂
We now know that the length from the end of the vertical axis to either focal point is a.

Now apply the Pythagorean Theorem to the triangle OB₁F₁, where O is the centre of the ellispe: a = B₁F₁; b = OB₁; c = OF₁	a² = b² + c²
Solve this equation for c, taking only the positive root:	c² = a² - b² ==> c = (a² - b²)^½
Finally, we have found an expression for c, the focal length. c = (a² - b²)^½