2. Circles & Ellipses (Conic Sections)

L3CALC

Author: Juyoung Park & Sedecember juyoung.parkk8@gmail.com & a3060100s@gmail.com

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Ellipses

An ellipse is defined as the set of all points in a plane where the sum of the distances from any point on the curve to two fixed points (called foci) is constant.

Since diagrams are way better than dictionary definitions, here’s the visual diagram.
Diagram link here
..Where $a$ is the point on the ellipse that crosses the $x$-axis, and $b$ is the point on the ellipse that crosses the $y$-axis.

P.S. I realized after making the diagram that $a$ is the distance not the coordinates sorry for the confusion..

The definition states that the sum of the distances from a point to both foci is constant. This means that in the diagram, the distance of the blue line + distance of the red line = distance of the purple line + distance of the green line.

Now, lets try to form the equation of the ellipse using this definition! We know that

$$\sqrt{(x-c{)}^2+y^2}+\sqrt{(x+c{)}^2+y^2}=(a-c)+(a+c)$$ $$\underset{\text{Sum of distances from foci to }(x,y)}{\underbrace{\sqrt{(x-c{)}^2+y^2}+\sqrt{(x+c{)}^2+y^2}}}=\underset{\text{Sum of distances from }(a,0)}{\underbrace{(a-c)+(a+c)}}$$

You can do any other point on the curve, but using the distance from $(a,0)$ makes the calculation easier since it is the $x$-intercept (the zero in the equation makes things easier 😁)

$$⟹\sqrt{(x-c{)}^2+y^2}+\sqrt{(x+c{)}^2+y^2}=2a$$

Now we can do some algebraic manipulation to simplify the equation.

$$\sqrt{(x-c{)}^2+y^2}=2a-\sqrt{(x+c{)}^2+y^2}$$ $$(x-c{)}^2+y^2=4a^2-4a\sqrt{(x+c{)}^2+y^2}+(x+c{)}^2+y^2$$ $$\cancel{x^2}-2cx+\cancel{c^2}+\cancel{y^2}=4a^2-4a\sqrt{(x+c{)}^2+y^2}+\cancel{x^2}+2cx+\cancel{c^2}+\cancel{y^2}$$ $$⟹-2cx=4a^2-4a\sqrt{(x+c{)}^2+y^2}+2cx$$ $$-4cx=4a^2-4a\sqrt{(x+c{)}^2+y^2}$$ $$-4cx-4a^2=-4a\sqrt{(x+c{)}^2+y^2}$$ $$cx+a^2=a\sqrt{(x+c{)}^2+y^2}$$ $${\left[cx+a^2=a\sqrt{(x+c{)}^2+y^2}\right]}^2$$ $$(a^2+cx{)}^2=a^2\left[(x+c{)}^2+y^2\right]$$ $$a^4+2a^{2}cx+c^2{x}^2=a^2\left[x^2+2cx+c^2+y^2\right]$$ $$a^4+\cancel{2a^{2}cx}+c^2{x}^2=a^2{x}^2+\cancel{2a^{2}cx}+a^2{c}^2+a^2{y}^2$$ $$a^4+c^2{x}^2-a^2{x}^2-a^2{c}^2-a^2{y}^2=0$$ $$(c^2-a^2)x^2-a^2{y}^2+a^4-a^2{c}^2=0$$

Now we can use the identity $c^2+b^2=a^2$. If you’re wondering where this identity came from, check the ellipse diagram then apply Pythagoras (or just check the formula sheet since it also says there)

$$⟹-b^2{x}^2-a^2{y}^2+b^2{a}^2=0$$

Now divide all sides by $a^2{b}^2$

$$⟹\boxed{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$$

..which is an equivalent expression to the one on the formula sheet!

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Circles

Circles are pretty much the same thing as ellipses except it’s a but squished.

In the above derived formula of ellipse

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

If $a\equiv b$ then it’s a circle. If $a\ne b$, then it’s an ellipse.

Do circles have a foci? Yes, a foci of a circle is the center of the circle.