Let us cut the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$ with the family of parallel lines

(1)

(2)

(3)

Theorem 1. The diameter of ellipse, i.e. the bisector of the parallel chords of an ellipse, is a line passing through the centre of the ellipse.

We see from the last equation, that the slope $m'$ of the diameter which bisects the chords with the slope $m$ is $m' = -\frac{b^2}{a^2m}$ ; accordingly one has the symmetric equation $$mm' = -\frac{b^2}{a^2}$$ between $m$ and $m'$ . From it one can infer that, conversely, the diameter with the slope $m$ bisects all chords which have the slope $m'$ . If we have two diameters of the ellipse, each one bisecting the chords parallel to the other, then these chords are conjugate diameters of each other. Apparently, the major axis and the minor axis are a pair of conjugate diameters.

If the line (1) especially is a tangent of the ellipse, then the points $P_1$ and $P_2$ and the midpoint $(x_0,\,y_0)$ coincide, and thus the equation (3) gives the slope of the tangent: $$m_t = -\frac{b^2x_0}{a^2y_0}$$ So we may write the equation of the tangent $$y-y_0 = -\frac{b^2x_0}{a^2y_0}(x-x_0)$$ of the ellipse. This can be simplified to the well-memorable form $$\frac{x_0x}{a^2}+\frac{y_0y}{b^2} = 1$$ where $(x_0,\,y_0)$ is the tangency point on the ellipse. One has also the

Theorem 2. The tangent of ellipse passing through an endpoint of a diameter is parallel to the conjugate diameter.

Notes. The sum of squares of any pair of conjugate radii is equal to $a^2+b^2$ . The ellipse has only one pair of equally long conjugate diameters, viz. the ones lying on the diagonals of the rectangle $x = \pm{a},\; y = \pm{b}$ . In the coordinate system $(\xi,\,\eta)$ with coordinate axes along a pair of conjugate radii $\alpha,\,\beta$ , the equation of the ellipse reads $\frac{\xi^2}{\alpha^2}+\frac{\eta^2}{\beta^2} = 1$ .

Bibliography

1

LAURI PIMIÄ: Analyyttinen geometria. Werner Söderström Osakeyhtiö, Porvoo and Helsinki (1958).