12.7 Ellipses: from definition to equation

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A teacher's guide to the connection between the definition and equation of an ellipse, and how to get from one to the other.
 Here is the geometric definition of an ellipse. There are two points called the “foci”: in this case, (-3,0) and (3,0) . A point is on the ellipse if the sum of its distances to both foci is a certain constant: in this case, I’ll use 10 . Note that the foci define the ellipse, but are not part of it.

The point ( $x$ , $y$ ) represents any point on the ellipse. $d1$ is its distance from the first focus, and $d2$ to the second. So the ellipse is defined geometrically by the relationship: $d1+d2=10$ .

To calculate $d1$ and $d2$ , we use the Pythagorean Theorem as always: drop a straight line down from ( $x$ , $y$ ) to create the right triangles. Please verify this result for yourself! You should find that $d1=\sqrt{\left(x+3{\right)}^{2}+{y}^{2}}$ and $d2=\sqrt{\left(x-3{\right)}^{2}+{y}^{2}}$ . So the equation becomes:

$\sqrt{\left(x+3{\right)}^{2}+{y}^{2}}+\sqrt{\left(x-3{\right)}^{2}+{y}^{2}}=10$ . This defines our ellipse

The goal now is to simplify it. We did problems like this earlier in the year (radical equations, the “harder” variety that have two radicals). The way you do it is by isolating the square root, and then squaring both sides. In this case, there are two square roots, so we will need to go through that process twice.

 $\sqrt{\left(x+3{\right)}^{2}+{y}^{2}}=10–\sqrt{\left(x-3{\right)}^{2}+{y}^{2}}$ Isolate a radical $\left(x+3{\right)}^{2}+{y}^{2}=100–20\sqrt{\left(x-3{\right)}^{2}+{y}^{2}}+\left(x–3{\right)}^{2}+{y}^{2}$ Square both sides $\left({x}^{2}+6x+9\right)+{y}^{2}=100–20\sqrt{\left(x-3{\right)}^{2}+{y}^{2}}+\left({x}^{2}–6x+9\right)+{y}^{2}$ Multiply out the squares $12x=100–20\sqrt{\left(x-3{\right)}^{2}+{y}^{2}}$ Cancel&combine like terms $\sqrt{\left(x-3{\right)}^{2}+{y}^{2}}=5–\frac{3}{5}x$ Rearrange, divide by 20 $\left(x–3{\right)}^{2}+{y}^{2}=25–6x+\frac{9}{\text{25}}{x}^{2}$ Square both sides again $\left({x}^{2}–6x+9\right)+{y}^{2}=25–6x+\frac{9}{\text{25}}{x}^{2}$ Multiply out the square $\frac{\text{16}}{\text{25}}{x}^{2}+{y}^{2}=16$ Combine like terms $\frac{{x}^{2}}{\text{25}}+\frac{{y}^{2}}{\text{16}}=1$ Divide by 16

…and we’re done! Now, according to the “machinery” of ellipses, what should that equation look like? Horizontal or vertical? Where should the center be? What are $a$ , $b$ , and $c$ ? Does all that match the picture we started with?

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