# Prove that a conic section is symmetrical with respect to its principal axis.

A Calculus book that I'm self-studying is asking me to prove the following theorem about conic sections:

A conic section is symmetrical with respect to its principal axis.

Here is my attempt at a solution:

I will use the following definition of a conic section:

A conic section is the set of all points $P$ such that

$|\overline{FP}|=e|\overline{RP}|$,

where $P$ is a point in a plane, $e$ is the eccentricity, $F$ is the focus of the conic section and $R$ is the point in the directrix such that the line $\overline{RP}$ is perpendicular to the directrix. In other words, $|\overline{RP}|$ is the distance from $P$ to the directrix.

I will also use the following definitions:

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$

The distance between a point $P(x,y)$ and the line $r: Ax+By+C=0$ is $|Ax+By+C| / \sqrt{A^2+B^2}$.

(Edit) The principal axis of a conic section is the line that goes through the focus $F$ and is perpendicular to the directrix.

For this solution, let's consider a conic section, and let $e$ be the eccentricity and $F$ be a focus of this conic section.

Let $d$ be the distance between the focus and the directrix. Let's place this conic section in a Cartesian coordinate system in such a way that:

1) the focus $F$ is in the origin, that is, $F$ is the point $(0,0)$;

2) the directrix is the line $x=-d$, or $x+d=0$.

In this particular configuration, the principal axis of the conic section is along the x-axis.

The value of $|\overline{FP}|$ is $\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$. The value of $|\overline{RP}|$ is the value of the distance between point $P$ and the line $x+d=0$. Using the formula $|Ax+By+C| / \sqrt{A^2+B^2}$ with $A=1$, $B=0$ and $C=d$, we get $|\overline{RP}|=|x+d|$. Substituting these values in the definition of a conic section:

$|\overline{FP}|=e|\overline{RP}|$

$\sqrt{x^2+y^2}=e|x+d|$

Squaring both sides:

$x^2+y^2=e^2(x+d)^2$

$x^2+y^2=e^2 (x^2+2xd+d^2)$

Or:

$y=\pm \sqrt{e^2 (x^2+2xd+d^2)-x^2}$

The above result means that the graph of $y$ (that is, the conic section) is symmetrical with respect to the x-axis, and, therefore, the conic section is symmetrical with respect to the principal its axis. Is this correct or am I missing something?

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Looks okay. You could stop once you reach the equation $\sqrt{x^2+y^2}=e|x+d|$, where it's clear that, for any point $(x,y)$ satisfying the equation, the point $(x,-y)$ will satisfy it, too. In fact, you could stop upon merely determining that $|FP| = \sqrt{x^2+y^2}$ and $|RP|=|x+d|$; the $|FP|$ condition is symmetric in $y$ (that is, if some "$y$" works, then "$-y$" works, too), and the $|RP|$ condition is completely independent of (and, hence, also symmetric in) $y$. –  Blue Jun 20 '12 at 18:35
(Missed the editing window!) One might say that $|FP|$ and $|RP|$ distances are "obviously" independent of $y$, even without giving the coordinatized formulations for them ... but, then, what's to prove? :) The point of the exercise appears to be to get you to recognize geometric symmetry in algebraic form: if all the "$y$"s in a relation are squared (or even-powered, or absolute-valued), then the graph is symmetric about the $x$-axis; likewise, even-powered "$x$"s imply symmetry about $y$. (Of course, you don't have the latter here, because there's a stray odd-powered $x$ floating around.) –  Blue Jun 20 '12 at 18:49

If there is some other definition of the principal axis in use, such as the $x$ axis when the ellipse has equation $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ with $|a| > |b|$, a separate argument would be needed to demonstrate that it coincides with the line of symmetry.