Conic Sections

TODO: Note the equations.

Parabola

A parabola is the set of all points that are equidistant from a point (the focus) and a fixed line (the directrix). If the focus is at \((0,p)\), and the directrix is \(y=-p\), then the equation of the parabola is:

\begin{equation*} x^{2}=4py \end{equation*}

If you rotate it, then it is:

\begin{equation*} y^{2}=4px \end{equation*}

with the focus being at \((p,0)\)

One of the properties of the parabola is that if you draw a straight line from the focus to a point on the parabola, then the “reflection” will be a horizontal line (i.e. the angle it makes with the tangent is the same as the angle a horizontal line will make with the tangent).

Showing the focus property of a parabola

The claim is that \(\alpha=\beta\). To show this, note that \(\beta\) is also the angle \(\angle PQF\). Then by one of the properties of triangles, the angle made by the line \(PF\) with the x-axis (call it \(\gamma\)) is \(\alpha+\beta\).

So what is \(\beta\)? Well, \(\tan\beta\) is the slope of the line \(PQ\), which we can get from calculus as \(PQ\) is the tangent line. Similarly, \(\tan\gamma\) can be obtained by calculating the slope of the \(PF\) line: We know the coordinates of \(P\) as we picked the point arbitrarily.

Finally, use the identity:

\begin{equation*} \tan\alpha=\tan(\gamma-\beta)=\frac{\tan\gamma-\tan\beta}{1+\tan\gamma\tan\beta} \end{equation*}

When you plug it all in, you’ll get \(\tan\alpha\) is the same as \(\tan\beta\) (the slope of the line \(PQ\)).

Below is Sage code that does this:

p, x, y, x0, y0, a, b, gamma = var(r"p x y x_0 y_0 alpha beta gamma")
m1 = 2*p/y0
print(f"Slope of PQ is: {m1}")
x0 = y0^2/(4*p)
m2 = y0/(x0-p)
print(f"Slope of PF is: {m2}")
C = (m2 - m1)/(1 + m1*m2)
print(f"tan(alpha) = {C.full_simplify()}")

Slope of PQ is: 2*p/y_0
Slope of PF is: -4*y_0/(4*p - y_0^2/p)
tan(alpha) = 2*p/y_0

Note that \(\tan\alpha\) is the same as the slope of \(PQ\) (\(\tan\beta\))

This is why parabolas are used for headlights, etc.

Any curve of the form \(y=ax^{2}+bx+c\) is a parabola. The vertex of that parabola occurs at \(x=-\frac{b}{2a}\)

Ellipse

An ellipse is the set of all points such that the sum of two distances to two fixed points (the foci) is constant.

Let the foci be on the x-axis with coordinates \((\pm c,0)\). Let the sum of the distances by \(a\). Then the formula of the ellipse is:

\begin{equation*} \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \end{equation*}

where \(b^{2}=a^{2}-c^{2}\).

Note that the sum of the distances to the two fixed points is \(2a\) (by construction).

Here, by design, \(a>b\) and of course \(a>c\).

An ellipse has the property that if we shoot a ray from one focus to the ellipse, its reflection will land on the other focus.

Proof of the reflective property of the ellipse.

To prove this, take a point \(P\) on the ellipse. Extend the line \(PF'\) until you hit a length of \(2a\). Call this point \(L\). Let \(w\) be the line that bisects the angle between \(PF\) and \(PF'\). We need to show that this line is the tangent. To do it, pick any point on this line and call it \(Q\). We have to show that \(Q\) cannot appear on the ellipse.

From the triangle inequality, we have:

\begin{equation*} 2 a=\left|L F'\right|<\left|Q F'\right|+|Q L|=\left|Q F'\right|+\left|Q F\right| \end{equation*}

Thus, \(Q\) cannot be on the ellipse, as for it to be so, the sum on the RHS must be exactly \(2a\).

Hyperbola

A hyperbola is the set of all points such that the difference of two distances to two fixed points (the foci) is constant. Let this difference be \(2a\), and let the foci be at \((\pm c,0)\)

Define \(b^{2}=c^{2}-a^{2}\) and we get this as the equation of a hyperbola:

\begin{equation*} \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \end{equation*}

It also has a reflective property: If we shoot a ray towards one of the foci, it will reflect and head to the other focus. The proof is similar to that of the ellipse, using this diagram:

Proof of the reflective property of the hyperbola