Focal Distance in the Context of an Ellipse

The focal distance in an ellipse refers to the distance from a point on the ellipse to one of the foci of the ellipse. An ellipse has two foci, and the sum of the distances from any point on the ellipse to the two foci is constant. This constant sum is equal to the major axis of the ellipse.

Definition and Formulas

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. If we denote the foci as $F_1$ and $F_2$, and a point on the ellipse as $P$, then the focal distance is the distance from $P$ to one of the foci, say $F_1$. We can denote this distance as $PF_1$.

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 $c = \sqrt{a^2 - b^2}$.

The focal distances for a point $(x, y)$ on the ellipse to the foci $F_1(-c, 0)$ and $F_2(c, 0)$ are given by:

$$PF_1 = \sqrt{(x + c)^2 + y^2}$$ $$PF_2 = \sqrt{(x - c)^2 + y^2}$$

The sum of these distances is constant and equal to $2a$:

$$PF_1 + PF_2 = 2a$$

Table of Differences and Important Points

Property	Description
Foci	Two fixed points inside the ellipse.
Semi-major axis (a)	Half the length of the longest diameter of the ellipse.
Semi-minor axis (b)	Half the length of the shortest diameter of the ellipse.
Focal distance (PF)	Distance from a point on the ellipse to one of the foci.
Constant sum	The sum of the focal distances from any point on the ellipse to both foci is always $2a$.
Eccentricity (e)	Defined as $e = \frac{c}{a}$, where $0 < e < 1$ for an ellipse.

Examples

Example 1: Finding the Focal Distances

Given an ellipse with the equation $\frac{x^2}{16} + \frac{y^2}{9} = 1$, find the focal distances to the point $(4, 0)$.

Solution:

First, we identify $a^2 = 16$ and $b^2 = 9$, so $a = 4$ and $b = 3$. We calculate $c$:

$$c = \sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7}$$

The foci are located at $(\pm\sqrt{7}, 0)$. Now we find the focal distances to the point $(4, 0)$:

$$PF_1 = \sqrt{(4 + \sqrt{7})^2 + 0^2} = 4 + \sqrt{7}$$ $$PF_2 = \sqrt{(4 - \sqrt{7})^2 + 0^2} = 4 - \sqrt{7}$$

The sum of these distances is:

$$PF_1 + PF_2 = (4 + \sqrt{7}) + (4 - \sqrt{7}) = 8$$

This sum is equal to $2a$, which is $2 \times 4 = 8$, confirming the property of the ellipse.

Example 2: Using the Focal Distances to Find a Point on the Ellipse

Given an ellipse with the equation $\frac{x^2}{25} + \frac{y^2}{16} = 1$ and a point $P$ on the ellipse such that $PF_1 = 9$, find the coordinates of $P$.

Solution:

First, we identify $a^2 = 25$ and $b^2 = 16$, so $a = 5$ and $b = 4$. We calculate $c$:

$$c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = \sqrt{9} = 3$$

The foci are located at $(\pm 3, 0)$. Since $PF_1 = 9$, and we know that $PF_1 + PF_2 = 2a = 10$, we can find $PF_2$:

$$PF_2 = 10 - PF_1 = 10 - 9 = 1$$

Now we have two equations:

$$PF_1 = \sqrt{(x + 3)^2 + y^2} = 9$$ $$PF_2 = \sqrt{(x - 3)^2 + y^2} = 1$$

Solving these equations simultaneously will give us the coordinates of $P$. However, since this is a more complex algebraic problem, we will not solve it here. But in general, you would square both sides of each equation, subtract one from the other to eliminate $y^2$, and solve for $x$. Then use the value of $x$ to find $y$.

Understanding focal distance and its properties is crucial for solving problems related to ellipses in mathematics, especially in geometry and calculus.