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# conjugate diameters of ellipse

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\;=\;1$ |

with the family of parallel lines

$\displaystyle y\;=\;mx\!+\!k$ | (1) |

having the common slope $m$; the number $k$ being a parameter. Substituting the expression (1) of $y$ to the equation of the ellipse, we get the quadratic equation

$\displaystyle(a^{2}m^{2}\!+\!b^{2})x^{2}+2a^{2}mkx+a^{2}(k^{2}\!-\!b^{2})\;=\;0$ | (2) |

determining the abscissas of the intersection points $P_{1}$ and $P_{2}$. The midpoint of the chord $P_{1}P_{2}$ of the ellipse is determined by the abscissa $x_{0}$, which is the arithmetic mean of the abscissas of $P_{1}$ and $P_{2}$, i.e. the roots of (2). Using the well-known properties of quadratic equation, we get

$x_{0}\;=\;-\frac{2a^{2}mk}{a^{2}m^{2}\!+\!b^{2}}\!:\!2\;=\;-\frac{a^{2}mk}{a^{% 2}m^{2}+b^{2}}.$ |

The ordinate $y_{0}$ of the midpoint of the chord satisfies

$y_{0}\;=\;mx_{0}\!+\!k.$ |

Eliminating the parameter $k$ from the two last equations gives us

$\displaystyle y_{0}\;=\;-\frac{b^{2}}{a^{2}m}x_{0}.$ | (3) |

This equation says that the point $(x_{0},\,y_{0})$ is situated on the line $y=-\frac{b^{2}}{a^{2}m}x$, which passes through the origin, the centre of the ellipse. So we have the

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 $m^{{\prime}}$ of the diameter which bisects the chords with the slope $m$ is $m^{{\prime}}=-\frac{b^{2}}{a^{2}m}$; accordingly one has the symmetric equation

$mm^{{\prime}}\;=\;-\frac{b^{2}}{a^{2}}$ |

between $m$ and $m^{{\prime}}$. From it one can infer that, conversely, the diameter with the slope $m$ bisects all chords which have the slope $m^{{\prime}}$. 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:

$m_{t}\;=\;-\frac{b^{2}x_{0}}{a^{2}y_{0}}$ |

So we may write the equation of the tangent

$y\!-\!y_{0}\;=\;-\frac{b^{2}x_{0}}{a^{2}y_{0}}(x\!-\!x_{0})$ |

of the ellipse. This can be simplified to the well-memorable form

$\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^{2}}\;=\;1$ |

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 $\frac{\xi^{2}}{\alpha^{2}}+\frac{\eta^{2}}{\beta^{2}}=1$.

# References

- 1 Lauri Pimiä: Analyyttinen geometria. Werner Söderström Osakeyhtiö, Porvoo and Helsinki (1958).

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## Comments

## Isn't It Conjugate vs. Transverse?

Given the two principal axes, x and y, isn't it just the

north-south y-axis is the conjugate axis, while the east-west, *perpendicular* x-axis is the transverse axis? See

http://books.google.com/books?id=Uk4wAAAAMAAJ&pg=RA1-PA381&zoom=3&sig=3Q...

http://books.google.com/books?id=OHsLAAAAYAAJ&pg=PA182&zoom=3&sig=_N9YWB...

~Kaimbridge~

## Re: Isn't It Conjugate vs. Transverse?

The transverse axis and the conjugate axis concern only the hyperbola; the former is the line segment connecting the vertices. See http://planetmath/encyclopedia/Hyperbola.html.

Jussi

## Re: Isn't It Conjugate vs. Transverse?

The transverse axis and the conjugate axis concern only the hyperbola; the former is the line segment connecting the vertices. See http://planetmath.org/encyclopedia/Hyperbola.html. (I corrected the URL.)

Jussi

## Re: Isn't It Conjugate vs. Transverse?

Excuse, the following is the _correct_ URL:

http://planetmath/encyclopedia/Hyperbola2.html.

Jussi

## Re: Isn't It Conjugate vs. Transverse?

Oh excuse still, it is

http://planetmath.org/encyclopedia/Hyperbola2.html.

## Re: Isn't It Conjugate vs. Transverse?

But what about the sources I noted? They both are talking about ellipses, not hyperbolae *AND* when talking about great circle delineation, you are dealing with a transverse graticule (i.e, the polar vertex is "pulled down" to the equator). See

http://www.fas.org/irp/agency/nima/nug/gloss_t.html

http://upload.wikimedia.org/wikipedia/en/5/5b/Graticule_Perspectives.PNG

~Kaimbridge~

## Re: Isn't It Conjugate vs. Transverse?

Dear Kaimbridge,

Maybe on your speciality (geodesy?), those terms are as you say. In mathematics, we speak only of the transverse axis and the conjugate axis of a hyperbola; see also in Wiki

http://en.wikipedia.org/wiki/Conjugate_axis.

Wolfram (http://mathworld.wolfram.com/) doesn't know tose terms.

Best regards,

Jussi

## Re: Isn't It Conjugate vs. Transverse?

P. S. -- As for ellipse, one speaks of major and minor (semi)axes.