Understanding the Focal Chord

In the context of conic sections, a focal chord is a line segment that passes through the focus of the conic and has both of its endpoints lying on the curve. For an ellipse, a focal chord is a line segment that passes through one of the foci and intersects the ellipse at two points.

Properties of a Focal Chord in an Ellipse

An ellipse is defined as the set of all points for which the sum of the distances to two fixed points (the foci) is constant. The standard equation of an ellipse with its center at the origin and its major axis along the x-axis is given by:

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

where $a$ is the semi-major axis and $b$ is the semi-minor axis. The foci are located at $(\pm c, 0)$, where $c = \sqrt{a^2 - b^2}$.

Table of Important Points

Property	Description
Focus	One of the two fixed points inside the ellipse. For a focal chord, it is the point through which the chord passes.
Endpoints	The points where the focal chord intersects the ellipse.
Length	The distance between the endpoints of the focal chord.
Direction	A focal chord can be perpendicular to the major axis or at any other angle, depending on the position of its endpoints.

Formulas Related to Focal Chords

For a focal chord of an ellipse that passes through the focus at $(c, 0)$, if the endpoints are $(x_1, y_1)$ and $(x_2, y_2)$, then the following properties hold:

1. The product of the distances of the endpoints from the center of the ellipse is equal to $b^2$: $$ (x_1 - 0)(x_2 - 0) = b^2 $$

2. The slope of the focal chord is given by: $$ m = \frac{y_2 - y_1}{x_2 - x_1} $$

3. The midpoint of the focal chord lies on the auxiliary circle of the ellipse, which has the equation: $$ x^2 + y^2 = a^2 $$

Examples

Example 1: Finding the Length of a Focal Chord

Consider an ellipse with the equation $\frac{x^2}{25} + \frac{y^2}{9} = 1$. Let's find the length of a focal chord that passes through the focus at $(4, 0)$.

1. First, we find the semi-major axis $a$ and semi-minor axis $b$: $$ a = 5, \quad b = 3 $$

2. The foci are located at $(\pm c, 0)$, where $c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = 4$.

3. Let the endpoints of the focal chord be $(x_1, y_1)$ and $(x_2, y_2)$. Since the chord passes through the focus at $(4, 0)$, one endpoint is $(4, y_1)$.

4. Using the property that the product of the distances from the center is $b^2$, we have: $$ (x_1 - 0)(x_2 - 0) = 9 $$ Since one endpoint is $(4, y_1)$, we have: $$ (4)(x_2) = 9 $$ $$ x_2 = \frac{9}{4} $$

5. To find $y_2$, we substitute $x_2$ into the equation of the ellipse: $$ \frac{(\frac{9}{4})^2}{25} + \frac{y_2^2}{9} = 1 $$ $$ \frac{81}{400} + \frac{y_2^2}{9} = 1 $$ $$ y_2^2 = 9(1 - \frac{81}{400}) $$ $$ y_2 = \pm 3\sqrt{1 - \frac{81}{400}} $$ $$ y_2 = \pm \frac{3}{20}\sqrt{319} $$

6. The length of the focal chord is the distance between the endpoints: $$ L = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ $$ L = \sqrt{(\frac{9}{4} - 4)^2 + (\frac{3}{20}\sqrt{319} - 0)^2} $$ $$ L = \sqrt{(\frac{1}{4})^2 + (\frac{3}{20}\sqrt{319})^2} $$ $$ L = \sqrt{\frac{1}{16} + \frac{9}{400}(319)} $$ $$ L = \sqrt{\frac{1}{16} + \frac{2871}{400}} $$ $$ L = \sqrt{\frac{25}{400} + \frac{2871}{400}} $$ $$ L = \sqrt{\frac{2896}{400}} $$ $$ L = \frac{54}{20} $$ $$ L = 2.7 $$

Therefore, the length of the focal chord is $2.7$ units.

Example 2: Focal Chord Perpendicular to the Major Axis

For a focal chord perpendicular to the major axis, the endpoints will have the same y-coordinate, and the chord will be a horizontal line segment. If the ellipse is the same as in Example 1, and the focal chord is perpendicular to the major axis at the focus $(4, 0)$, then the endpoints will be of the form $(4 \pm d, 0)$, where $d$ is half the length of the focal chord.

Since the chord is perpendicular to the major axis, it will be parallel to the minor axis, and its length can be found by using the property of the ellipse that relates the distance from the center to the endpoints of the chord:

$$ (4 + d)(4 - d) = b^2 $$ $$ 16 - d^2 = 9 $$ $$ d^2 = 7 $$ $$ d = \sqrt{7} $$

The length of the focal chord is then $2d = 2\sqrt{7}$ units.

In conclusion, focal chords in an ellipse have unique properties that relate to the geometry of the ellipse itself. By understanding these properties and applying the relevant formulas, one can solve problems involving focal chords in an ellipse.