# Simple proofs of the reflection properties of conics using calculus

An ellipse can be described as the locus of points whose sum of distances from two given points, the foci, is constant. Thus, we can write the equation as: \[ {\left|{\mathbf{x}}- {\mathbf{a}}\right|} + {\left|{\mathbf{x}}- {\mathbf{b}}\right|} = k \] where \({\mathbf{a}}, {\mathbf{b}}\) are the foci of the ellipse, and \(k\) is some positive constant. Now, if we imagine that \({\mathbf{x}}= {\mathbf{x}}(t)\) is some parametrization of the ellipse, we can find an equation satisfied by the tangent vector \({\mathbf{x}}'\) simply by taking derivatives on both sides (and rearranging): \[ {\mathbf{x}}'\cdot\frac{{\mathbf{x}}-{\mathbf{a}}}{{\left|{\mathbf{x}}-{\mathbf{a}}\right|}} + {\mathbf{x}}'\cdot\frac{{\mathbf{x}}-{\mathbf{b}}}{{\left|{\mathbf{x}}-{\mathbf{b}}\right|}} = 0. \] Since \(\frac{{\mathbf{x}}-{\mathbf{a}}}{{\left|{\mathbf{x}}-{\mathbf{a}}\right|}}\) is the unit vector pointing from \({\mathbf{x}}\) to the focus \({\mathbf{a}}\), and similarly for \({\mathbf{b}}\), this equation says that the tangent vector \({\mathbf{x}}'\) makes complementary angles with the lines from \({\mathbf{x}}\) to the foci. (If the cosines of two angles add up to zero, the angles add up to \(180^\circ\).)

Notice that in the special case of the circle (i.e., \({\mathbf{a}}= {\mathbf{b}}\)), the equation simplifies to \({\mathbf{x}}' \cdot \frac{{\mathbf{x}}-{\mathbf{a}}}{{\left|{\mathbf{x}}-{\mathbf{a}}\right|}} = 0\), which says that the tangent line at a point on the circle is perpendicular to the line joining that point and the center of the circle.

The case of a hyperbola is very similar to the case of the ellipse: the
only difference is that now the *difference* of the two distances is
constant, rather than the sum, so the formulas have a minus sign instead
of plus. So the tangent vector instead of making complementary angles
with the lines from \({\mathbf{x}}\) to the foci, now makes equal angles.

A parabola can be described as the locus of points equidistant from a given point, the focus, and a given line, the directrix. To write down the equation, let’s assume the directrix passes through the origin of the plane and has normal vector \({\mathbf{n}}\). Then the equation is \[ {\mathbf{x}}\cdot {\mathbf{n}}= {\left|{\mathbf{x}}- {\mathbf{a}}\right|}, \] where \({\mathbf{a}}\) is the focus. Taking derivatives we get that \[ {\mathbf{x}}' \cdot {\mathbf{n}}= {\mathbf{x}}' \cdot \frac{{\mathbf{x}}-{\mathbf{a}}}{{\left|{\mathbf{x}}-{\mathbf{a}}\right|}}. \] This says that the angle between the tangent vector at \({\mathbf{x}}\) makes equal angles with the normal vector and the line joining \({\mathbf{x}}\) with the focus, \({\mathbf{a}}\).