Center of Conic is Center of Symmetry

Theorem

Let $\KK$ be a central conic.

Let $C$ be the center of $\KK$.

Then $C$ is a center of symmetry for $\KK$.

Proof

First we note from Parabola has No Center that we do not need to consider the parabola.

Then we note from Circle is Ellipse with Equal Major and Minor Axes that a circle is a special case of the ellipse.

Hence there is no need to investigate the circle separately.

It remains to demonstrate the result for the ellipse and the hyperbola.

Let the central conic be expressed in reduced form.

By definition of reduced form, the center of $\KK$ lies at the origin.

From Equation of Ellipse in Reduced Form and Equation of Hyperbola in Reduced Form, $\KK$ can be expressed as:

$\dfrac {x^2} {a^2} \pm \dfrac {y^2} {b^2} = 1$

Let $\tuple {p, q}$ be a point on $\KK$.

Then:

\(\ds \dfrac {p^2} {a^2} \pm \dfrac {q^2} {b^2}\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {\paren {-p}^2} {a^2} \pm \dfrac {\paren {-q}^2} {b^2}\)

\(=\)

\(\ds 1\)

as $p^2 = \paren {-p}^2$ and $q^2 = \paren {-q}^2$

and so $\tuple {-p, -q}$ is also a point on $\KK$.

Hence the result by definition of center of symmetry.

$\blacksquare$