Editing Ellipse (section)

=== Tangent ===
An arbitrary line <math>g</math> intersects an ellipse at 0, 1, or 2 points, respectively called an ''exterior line'', ''tangent'' and ''secant''. Through any point of an ellipse there is a unique tangent. The tangent at a point <math>(x_1,\, y_1)</math> of the ellipse <math>\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1</math> has the coordinate equation:
<math display="block">\frac{x_1}{a^2}x + \frac{y_1}{b^2}y = 1.</math>

A vector [[parametric equation]] of the tangent is:
<math display="block">\vec x = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + s \left(\begin{array}{r}
-y_1 a^2 \\
 x_1 b^2
\end{array}\right) , \quad s \in \R. </math>

'''Proof:'''
Let <math>(x_1,\, y_1)</math> be a point on an ellipse and <math display="inline">\vec{x} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + s \begin{pmatrix} u \\ v \end{pmatrix}</math> be the equation of any line <math>g</math> containing <math>(x_1,\, y_1)</math>. Inserting the line's equation into the ellipse equation and respecting <math display="inline">\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1</math> yields:
<math display="block">
  \frac{\left(x_1 + su\right)^2}{a^2} + \frac{\left(y_1 + sv\right)^2}{b^2} = 1\ \quad\Longrightarrow\quad
  2s\left(\frac{x_1u}{a^2} + \frac{y_1v}{b^2}\right) + s^2\left(\frac{u^2}{a^2} + \frac{v^2}{b^2}\right) = 0\ .</math>
There are then cases:
# <math>\frac{x_1}{a^2}u + \frac{y_1}{b^2}v = 0.</math> Then line <math>g</math> and the ellipse have only point <math>(x_1,\, y_1)</math> in common, and <math>g</math> is a tangent. The tangent direction has [[normal (geometry)|perpendicular vector]] <math>\begin{pmatrix} \frac{x_1}{a^2} & \frac{y_1}{b^2} \end{pmatrix}</math>, so the tangent line has equation <math display="inline">\frac{x_1}{a^2}x + \tfrac{y_1}{b^2}y = k</math> for some <math>k</math>. Because <math>(x_1,\, y_1)</math> is on the tangent and the ellipse, one obtains <math>k = 1</math>.
# <math>\frac{x_ 1}{a^2}u + \frac{y_1}{b^2}v \ne 0.</math> Then line <math>g</math> has a second point in common with the ellipse, and is a secant.

Using (1) one finds that <math>\begin{pmatrix} -y_1 a^2 & x_1 b^2 \end{pmatrix}</math> is a tangent vector at point <math>(x_1,\, y_1)</math>, which proves the vector equation.

If <math>(x_1, y_1)</math> and <math>(u, v)</math> are two points of the ellipse such that <math display="inline">\frac{x_1u}{a^2} + \tfrac{y_1v}{b^2} = 0</math>, then the points lie on two ''conjugate diameters'' (see [[#Conjugate diameters|below]]). (If <math>a = b</math>, the ellipse is a circle and "conjugate" means "orthogonal".)