Proving the formula for focal length of an ellipse
The narrator calls P and Q the lengths of the major and minor axes, but they are actually the semi-axes (centre to end). He partly corrects this around 90s; a note at 99.5s clarifies it for students.
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The video proves the relationship between an ellipse's semi-major length P, semi-minor length Q and focal length F. It finds the constant sum of distances to the foci in two ways: at the end of the major axis (2P) and at the end of the minor axis (2 times sqrt(F^2 + Q^2)). Setting these equal gives F^2 + Q^2 = P^2, so F = sqrt(P^2 - Q^2).
You'll be able to
- explain why the sum of distances to the foci equals 2P at the end of the major axis.
- use the Pythagorean theorem to find the distance from the end of the minor axis to a focus.
- derive and use F = sqrt(P^2 - Q^2) to find the focal length of an ellipse.
Key words
- ellipse
- The set of all points where the sum of the distances to two fixed points (the foci) is constant.
- foci
- The two fixed points inside an ellipse used to define it; one of them is called a focus.
- focal length
- The distance from the centre of the ellipse to one focus, called F in this video.
- major axis
- The longest line through the centre of the ellipse; in the video P is the distance from the centre to its end.
- minor axis
- The shortest line through the centre of the ellipse; in the video Q is the distance from the centre to its end.
- hypotenuse
- The longest side of a right triangle, opposite the right angle.
Transcript
- [Instructor] What I wanna do in this video is prove the mathematical relationship between the lengths of the major axis, which here I am calling P, the length of the minor axis of an ellipse, which I'm calling Q. And the focal length, which I am calling F right over here.
So we know that an ellipse is defined by the set of all points where the sum of the distances to the two foci is constant. So first of all, let's think about what that sum of that distance might be. Or let's think about two ways that that sum of distance, how we might express it.
And since it's constant, we can set them equal to each other. First, we can think about what happens when we are at a point like that. How far, what is the sum of the distance from the two foci? Well, the distance from this first foci right over there, that is going to be P minus F.
So let me just write in this color. So that's going to be P minus F. And then plus we want to add this distance right over here, which is going to be P plus F. So if you're at one of these extremes all the way out on your major axis, the sum of the distances to the two foci, the negative F and the F, they get canceled out.
And so, we're left with just two P, two times the length of the major axis. Or depending on how you define the major axis, two times the length from the origin to one of these extremes along the major axis. Now, we can do a similar exercise with one of the extremes along the minor axis.
Let's take this point right over there. What is the sum of the distances to the two foci? Well, here we can use the Pythagorean theorem. Right over here, we could set up two right triangles. That's a right triangle, that's a right triangle. We know what F and Q are, so this is the hypotonus.
So this is going to be F squared plus Q squared. The square root of that is going to be the length of an hypotonus. And the same thing is this one over here, it's going to be the square root of F squared plus Q squared. So the sum of the distances is going to be two times the square root of F squared plus Q squared.
Now, we just said that at any point along the ellipse, the sum of the distances to the two foci have to be constant. So this expression needs to be the same as this expression right over there. And, so we can divide both sides by two and then we can square both sides.
We would get F squared plus Q squared is equal to P squared. And so, there you have it. We already have a relationship between the focal length, the minor axis length, and the major axis length. But if we wanna solve for focal length, we could just subtract Q squared from both sides.
We get P squared minus Q squared, and we get the focal length, if we take the principle root of both sides, is equal to the square root of P squared minus Q squared. And we're done.
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