Video: Deriving the Equation of an Ellipse given Its Foci and the Length of Its Major Axis

Derive the equation of an ellipse with foci at the points (1, 3) and (6, 3) which has a major axis of length 15.

08:21

Video Transcript

Derive the equation of an ellipse with foci at the points one, three and six, three which has a major axis of length 15.

Let’s first try to sketch the ellipse before finding its equation. We start by setting up the coordinate plane. We mark in the two foci: the point one, three, which we’ll call 𝐹 one, and the point six, three, which we’ll call 𝐹 two. We also mark in the major axis of the ellipse, which we’re told has a length of 15 units.

Using this major axis as a guide, we can have a go at sketching what this ellipse might look like. It’s not important to be particularly accurate at this point. The major axis meets the ellipse at the two vertices, 𝑉 one and 𝑉 two, and we can use the fact that the major axis has a length of 15 units to find the coordinates of these two vertices.

We first find the coordinates of the point 𝐢, which is the center of the ellipse. This centre of the ellipses is not only the midpoint of the vertices of the ellipse, it’s also the midpoint of the two foci. And so using this fact and the midpoint formula, we can see that it’s coordinates are 3.5, three. Now as the center 𝐢 is the midpoint of the major axis, which has length 15, one end of the major axis, 𝑉 two, must be 7.5 units away and hence it has coordinates 11, three.

The same as of course true of 𝑉 one; it’s 7.5 units away, this time in the negative π‘₯-direction. And so it’s coordinates are negative four, three. You might know that there is an equation for an ellipse whose major axis is parallel to the π‘₯-axis and that this equation involves β„Ž and π‘˜, which are the coordinates of the center which we have found. It also involves π‘Ž, which is half the length of the major axis again that we have found, and 𝑏, which is half the length of the minor axis which we haven’t found yet but could easily find by using the Pythagorean theorem.

However, instead of using this general formula, we’re going to derive the equation of the ellipse from the first principles using the definition of an ellipse. So let’s remove some of the working that we don’t need and get on with this. Now we’re back with a much cleaner diagram. We just have the ellipse, its vertices and foci marked. Recall that the definition of an ellipse is the locus of points for which the sum of the distances from the foci is constant.

So taking a general point 𝑃 on our ellipse we know that 𝐹 one 𝑃, the distance from 𝐹 one to 𝑃, plus 𝐹 two 𝑃, the distance from 𝐹 two to 𝑃, is equal to some constant 𝐾. What is this constant 𝐾? Well we can work it out because this holds true for any point 𝑃 on the ellipse, including the two vertices 𝑉 one and 𝑉 two. So for the sake of argument, picking 𝑉 two, we can see that the distance from 𝐹 one to 𝑉 two plus the distance from 𝐹 two to 𝑉 two must be this constant 𝐾.

And looking at the coordinates of 𝐹 one and 𝑉 two, we can see that the distance between them is 11 minus one, which is 10. And now we can do the same for the 𝐹 two, 𝑉 two. Looking at the coordinates of 𝐹 two and 𝑉 two, we can see the distance between them is 11 minus six, which is five. And so on the left-hand side, we have 10 plus five, which is 15, and so 𝐾 is 15.

We can substitute this value in to show that for any point 𝑃 on the ellipse, the distance from 𝐹 one to 𝑃 plus the distance from 𝐹 two to 𝑃 is 15. We can now get rid of some of our working and focus on the equation 𝐹 one 𝑃 plus 𝐹 two 𝑃 is equal to 15. This is a very nice equation for our ellipse, but we’d really like one in terms of the coordinates of 𝑃, which are π‘₯ and 𝑦, instead. How we going to rewrite this distance from 𝐹 one to 𝑃 in terms of π‘₯ and 𝑦? Well, using the distance formula.

The distance from π‘₯ one, 𝑦 one to π‘₯ two, 𝑦 two is the square root of π‘₯ one minus π‘₯ two squared plus 𝑦 one minus 𝑦 two squared. So looking at the coordinates of 𝐹 one and 𝑃, we can see the distance from 𝐹 one to 𝑃 is the square root of one minus π‘₯ squared plus three minus 𝑦 squared. And we do something similar to find the distance from 𝐹 two to 𝑃. Using these expressions for 𝐹 one 𝑃 and 𝐹 two 𝑃, we have an equation for the ellipse in terms of π‘₯ and 𝑦, and so we don’t need our diagram anymore; it has served us well, but it has fulfilled its purpose.

The rest of this video will just be about manipulating the equation that we have to make it simpler. Okay, so let’s write the equation of the ellipse using these two expressions that we’ve got. Substituting these expressions in, we get that the square root of one minus π‘₯ squared plus three minus 𝑦 squared plus the square root of six minus π‘₯ squared plus three minus 𝑦 squared is equal to 15. Let’s see if we can get a simpler form of this equation which doesn’t involve square roots. So we first square both sides.

On the left-hand side, we get the square of the first term, which nicely gets rid of the square root sign, plus two times the product of the cross terms, which unfortunately doesn’t, plus the square of the second term; and on the right-hand side, we get 15 squared, which is 225. We can see two terms that we can combine, and I’m also going to rearrange the terms so that only the term involving the square roots is on the right-hand side.

And now that the square roots are isolated on the right-hand side, we can square again without getting any cross terms. On the left-hand side, we get a complete mess; on the right-hand side, the main effect is to remove the square roots. Of course, we can also now expand these brackets. And having done so, we can notice that there are some terms on the right which are also on the left, and so we can cancel. Here’s one; here’s another; and finally, this term also cancels out; and finally we notice a pair of like terms which don’t cancel but can be combined.

So tidying up and moving all the terms left-hand side, we get this expression is equal to zero. It turns out that if we expand and simplify all the terms involving π‘₯, we get something relatively straightforward: negative 800π‘₯ squared plus 5600π‘₯ minus 15425. And completing the square on this quadratic in π‘₯, we get negative 200 times two π‘₯ minus seven squared minus 5625. So we use this completed square form and bring down the other terms to get this equation.

Combining the two constant terms, rearranging so that they are on the right hand side and then multiplying both sides by negative one, we get 200 times two π‘₯ minus seven squared plus 900 times three minus 𝑦 squared equals 45000. And dividing both sides by 45000 so that the right-hand side is just one as is convention, we get two π‘₯ minus seven squared over 225 plus 𝑦 minus three squared over 50 equals one.

Note also that we rewrote three minus 𝑦 squared as 𝑦 minus three squared to put the variable first again as is convention. So it’s taken a lot of algebraic manipulation, but we’ve ended up with an equation which is quite simple and which doesn’t contain any square roots.

And having gone through this arduous process, we probably want to make sure that we don’t have to do it again. It therefore makes sense to derive a general equation for an ellipse that we can just substitute values into rather than having to go through this process every time you want to find the equation of an ellipse.

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