Understanding the Ellipse: Focus, Directrix, Vertex, and Latus Rectum

An ellipse is a set of points in a plane such that the sum of the distances from two fixed points (foci) is constant. It is a conic section that can be thought of as a stretched circle. To solve problems involving ellipses, it is crucial to understand its key elements: focus, directrix, vertex, and latus rectum.

Key Elements of an Ellipse

Element	Description
Focus (Foci)	Two fixed points inside the ellipse. The sum of the distances from any point on the ellipse to the foci is constant.
Directrix	A fixed line outside the ellipse. The ratio of the distance of any point on the ellipse from a focus to its distance from the corresponding directrix is constant (eccentricity).
Vertex	The points where the ellipse is widest (major vertices) or narrowest (minor vertices).
Latus Rectum	A line segment perpendicular to the major axis at a focus, with endpoints on the ellipse. Its length is directly related to the distance between the foci.

Standard Form of an Ellipse

The standard form of an ellipse with its center at the origin (0, 0) is given by:

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

where:

• $a$ is the semi-major axis length (distance from the center to a vertex on the major axis).
• $b$ is the semi-minor axis length (distance from the center to a vertex on the minor axis).

Important Formulas

• Eccentricity ($e$): $e = \sqrt{1 - \frac{b^2}{a^2}}$ for $a > b$.
• Focus (Foci): $(\pm ae, 0)$ for a horizontal ellipse, or $(0, \pm ae)$ for a vertical ellipse.
• Directrix: $x = \pm \frac{a}{e}$ for a horizontal ellipse, or $y = \pm \frac{a}{e}$ for a vertical ellipse.
• Vertex (Vertices): $(\pm a, 0)$ for a horizontal ellipse, or $(0, \pm a)$ for a vertical ellipse.
• Latus Rectum Length ($L$): $L = \frac{2b^2}{a}$ for a horizontal ellipse, or $L = \frac{2a^2}{b}$ for a vertical ellipse.

Examples

Example 1: Finding the Focus and Directrix

Given the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$, find the foci and directrices.

Solution:

1. Identify $a^2$ and $b^2$. Here, $a^2 = 16$ and $b^2 = 9$.
2. Calculate $a$ and $b$. $a = 4$ and $b = 3$.
3. Calculate the eccentricity $e$. $$ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} $$
4. Find the foci using the eccentricity. $$ Foci = (\pm ae, 0) = (\pm 4 \cdot \frac{\sqrt{7}}{4}, 0) = (\pm \sqrt{7}, 0) $$
5. Find the directrices. $$ Directrix = x = \pm \frac{a}{e} = \pm \frac{4}{\frac{\sqrt{7}}{4}} = \pm \frac{16}{\sqrt{7}} $$

Example 2: Finding the Latus Rectum

Given the same ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$, find the length of the latus rectum.

Solution:

1. Use the formula for the latus rectum length for a horizontal ellipse. $$ L = \frac{2b^2}{a} = \frac{2 \cdot 9}{4} = \frac{18}{4} = 4.5 $$
2. The length of the latus rectum is 4.5 units.

By understanding these elements and formulas, you can solve a wide range of problems related to ellipses. Practice with different equations of ellipses to become proficient in identifying and calculating the focus, directrix, vertices, and latus rectum.