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# properties of ellipse

The ellipse is defined as the locus of the points $P$ of the plane such that the distances of $P$ from two fixed points (the foci) have a constant sum ($2a$). The ellipse gets the simplest equation, when the foci are on a coordinate axis equally distant ($=c<a$) from the origin. So, if the foci are $(\pm c,\,0)$, the equation of the ellipse may be written first

$\displaystyle\sqrt{(x\!-\!c)^{2}\!+\!y^{2}}+\sqrt{(x\!+\!c)^{2}\!+\!y^{2}}\;=% \;2a.$ | (1) |

After two squarings it is simplified to

$\displaystyle(a^{2}\!-\!c^{2})x^{2}\!+\!a^{2}y^{2}\,=\,a^{2}(a^{2}\!-\!c^{2}).$ | (2) |

Denoting $a^{2}\!-\!c^{2}:=b^{2}$ (where $b>0$) the equation of the ellipse attains the form

$\displaystyle\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\;=\;1.$ | (3) |

From the equation one sees that $x=\pm a$ give the rightmost and the leftmost points of the ellipse (on the $x$-axis); the segment between them is the major axis, with length $2a$. Similarly, $y=\pm b$ give the highest and the lowest points (on the $y$-axis); the segment between them is the minor axis, with length $2b$. The distance between a focus and an end point of the minor axis is $a$ (see the yellow triangle).

The eccentricity $\varepsilon$ is the ratio of the “focal length” $2c$ and the major axis $2a$. The focal radii $r_{1}$ and $r_{2}$ satisfy

$r_{1}^{2}=(c-x)^{2}+y^{2},\quad r_{2}^{2}=(-c-x)^{2}+y^{2},$ |

whence the subtraction yields

$r_{2}^{2}\!-\!r_{1}^{2}\;=\;4cx.$ |

This gives, since $r_{2}\!+\!r_{1}=2a$, that

$r_{2}\!-\!r_{1}\;=\;2\varepsilon x,$ |

and the both last equations imply the expressions

$\displaystyle r_{1}\;=\;a\!-\!\varepsilon x,\quad r_{2}\;=\;a\!+\!\varepsilon x$ | (4) |

By the entry conjugate diameters of ellipse, the slope of the tangent of the ellipse in the point $(x_{0},\,y_{0})$ is

$\displaystyle m_{t}\;=\;-\frac{b^{2}x_{0}}{a^{2}y_{0}}.$ | (5) |

This result is easily obtained also by implicit differentiation of (3):

$\frac{2x}{a^{2}}+\frac{2yy^{{\prime}}}{b^{2}}\;=\;0$ |

Substituting here $x=x_{0}$, $y=y_{0}$ and solving $m_{t}=y^{{\prime}}$, one gets (5). Then we can write the equation of the normal line of the ellipse (3) in $(x_{0},\,y_{0})$:

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

The normal cuts the $x$-axis in the point $N$ with

$x\;=\;\frac{a^{2}-b^{2}}{a^{2}}x_{0}\;=\;\frac{c^{2}}{a^{2}}x_{0}\;=\;% \varepsilon^{2}x_{0}.$ |

Thus the distances of $N$ from the foci are by (4) equal to

$d_{1}\;=\;c-\varepsilon^{2}x_{0}=a\varepsilon-\varepsilon^{2}x_{0}=\varepsilon% (a-\varepsilon x_{0})=\varepsilon r_{1},$ |

$d_{2}\;=\;c+\varepsilon^{2}x_{0}=a\varepsilon+\varepsilon^{2}x_{0}=\varepsilon% (a+\varepsilon x_{0})=\varepsilon r_{2}.$ |

Consequently, $d_{1}\!:\!d_{2}\,=\,r_{1}\!:\!r_{2}$, which means by the bisectors theorem that the normal is the angle bisector of the angle between $r_{1}$ and $r_{2}$.

Accordingly, we have derived the

Theorem 1. The normal of ellipse bisects the angle between the focal radii. This may also be stated so that the tangent of the ellipse forms equal angles with both focal radii.

Corollary. If the circumference of an ellipse reflects the light rays meeting it, then the rays emanating from one of the foci meet after the reflection in the other focus.

One can prove also the

Theorem 2. The locus of the intersection point of a normal of an ellipse and the line through a focus and perpendicular to the normal is the circumscribed circle of the ellipse.

## Mathematics Subject Classification

53A04*no label found*51N20

*no label found*51-00

*no label found*

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