# 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

## 1 Answer

The question, at the time I write this, does not define "principal axis".

In the definitions of a conic using distances to two foci, or to a focus and a directrix, there is a natural line of symmetry where reflection across that line preserves the distances and is therefore a symmetry of the conic. This is the line through the two foci, or the line through the focus and perpendicular to the directrix. This line also happens to be the principal axis, usually by definition.

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.

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Thank you for the observation, I edited the original post to include the intended definition of principal axis. – anonymous Jun 20 '12 at 21:16