Ellipses

📌 What is an Ellipse?

An ellipse is the set of all points in a plane such that the sum of the distances from two fixed points (called foci) to any point on the ellipse is constant. An ellipse looks like a stretched or flattened circle.

Standard Form Equations of an Ellipse

Horizontal Major Axis:

\( \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \) where \( a > b \)

Center: \( (h, k) \)
Major axis is horizontal (left-right)
Vertices: \( (h \pm a, k) \)
Co-vertices: \( (h, k \pm b) \)
Foci: \( (h \pm c, k) \) where \( c^2 = a^2 - b^2 \)

Vertical Major Axis:

\( \frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1 \) where \( a > b \)

Center: \( (h, k) \)
Major axis is vertical (up-down)
Vertices: \( (h, k \pm a) \)
Co-vertices: \( (h \pm b, k) \)
Foci: \( (h, k \pm c) \) where \( c^2 = a^2 - b^2 \)

Key Rule - Identifying Major Axis:

⚠️ Important: \( a \) is always the larger value!

If the larger denominator is under \( x \)-term → Horizontal major axis
If the larger denominator is under \( y \)-term → Vertical major axis
\( a \) = semi-major axis (half the major axis length)
\( b \) = semi-minor axis (half the minor axis length)

Finding Center, Vertices, and Co-vertices

From Standard Form:

Find the center \( (h, k) \) directly from the equation
Identify \( a \) and \( b \): Take square roots of denominators
Determine which is larger to find the major axis direction
Calculate vertices by moving \( a \) units along the major axis
Calculate co-vertices by moving \( b \) units along the minor axis

📝 Example 1 - Horizontal Major Axis:

Find center, vertices, and co-vertices of \( \frac{(x - 3)^2}{25} + \frac{(y + 2)^2}{9} = 1 \)

Step 1: Center: \( (3, -2) \)
Step 2: \( a^2 = 25 \), so \( a = 5 \); \( b^2 = 9 \), so \( b = 3 \)
Step 3: Since 25 > 9 and 25 is under \( x \)-term, major axis is horizontal
Step 4: Vertices (move 5 units left/right from center):
\( (3 + 5, -2) = (8, -2) \) and \( (3 - 5, -2) = (-2, -2) \)
Step 5: Co-vertices (move 3 units up/down from center):
\( (3, -2 + 3) = (3, 1) \) and \( (3, -2 - 3) = (3, -5) \)

📝 Example 2 - Vertical Major Axis:

Find center, vertices, and co-vertices of \( \frac{(x + 1)^2}{16} + \frac{(y - 4)^2}{36} = 1 \)

Center: \( (-1, 4) \)
\( a^2 = 36 \), so \( a = 6 \); \( b^2 = 16 \), so \( b = 4 \)
Since 36 > 16 and 36 is under \( y \)-term, major axis is vertical
Vertices: \( (-1, 4 + 6) = (-1, 10) \) and \( (-1, 4 - 6) = (-1, -2) \)
Co-vertices: \( (-1 + 4, 4) = (3, 4) \) and \( (-1 - 4, 4) = (-5, 4) \)

Length of Major and Minor Axes

Formulas:

Length of Major Axis:

\( 2a \)

The distance between the two vertices

Length of Minor Axis:

\( 2b \)

The distance between the two co-vertices

📝 Examples - Axis Lengths:

Example 1: \( \frac{x^2}{49} + \frac{y^2}{25} = 1 \)

\( a^2 = 49 \), so \( a = 7 \); \( b^2 = 25 \), so \( b = 5 \)
Major axis length: \( 2a = 2(7) = 14 \)
Minor axis length: \( 2b = 2(5) = 10 \)

Example 2: \( \frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{16} = 1 \)

\( a = 4 \) (larger), \( b = 3 \) (smaller)
Major axis length: \( 2(4) = 8 \)
Minor axis length: \( 2(3) = 6 \)

Finding the Foci of an Ellipse

The Relationship Formula:

\( c^2 = a^2 - b^2 \)

Where:

\( c \) = distance from center to each focus
\( a \) = semi-major axis (larger value)
\( b \) = semi-minor axis (smaller value)
Foci are always on the major axis

Steps to Find Foci:

Identify \( a \) and \( b \) from the equation
Calculate \( c \) using \( c^2 = a^2 - b^2 \)
Determine major axis direction (horizontal or vertical)
Move \( c \) units from center along the major axis

📝 Example 1 - Horizontal:

Find the foci of \( \frac{(x - 1)^2}{36} + \frac{(y + 3)^2}{20} = 1 \)

Step 1: Center: \( (1, -3) \)
\( a^2 = 36 \), so \( a = 6 \); \( b^2 = 20 \), so \( b = \sqrt{20} = 2\sqrt{5} \)
Step 2: Calculate \( c \):
\( c^2 = 36 - 20 = 16 \), so \( c = 4 \)
Step 3: Major axis is horizontal (36 is under \( x \)-term)
Step 4: Foci are at:
\( (1 + 4, -3) = (5, -3) \) and \( (1 - 4, -3) = (-3, -3) \)

📝 Example 2 - Vertical:

Find the foci of \( \frac{x^2}{9} + \frac{y^2}{25} = 1 \)

Center: \( (0, 0) \)
\( a = 5, b = 3 \) (major axis is vertical)
\( c^2 = 25 - 9 = 16 \), so \( c = 4 \)
Foci: \( (0, 4) \) and \( (0, -4) \)

Writing Equations from Graphs

Steps:

Identify the center \( (h, k) \) from the graph
Find \( a \) by counting units from center to a vertex
Find \( b \) by counting units from center to a co-vertex
Determine orientation (horizontal or vertical major axis)
Write equation with larger value under the term matching major axis direction

📝 Example - From Graph:

An ellipse has center at \( (2, -1) \), vertices at \( (2, 4) \) and \( (2, -6) \), and co-vertices at \( (5, -1) \) and \( (-1, -1) \). Write the equation.

Step 1: Center: \( (2, -1) \)
Step 2: Distance from center to vertex:
\( a = |4 - (-1)| = 5 \)
Step 3: Distance from center to co-vertex:
\( b = |5 - 2| = 3 \)
Step 4: Major axis is vertical (vertices aligned vertically)
Equation: \( \frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{25} = 1 \)

Writing Equations Using Properties

Given Vertices and Co-vertices:

Find center using midpoint of vertices
Calculate \( a \) (distance from center to vertex)
Calculate \( b \) (distance from center to co-vertex)
Write equation in standard form

Given Foci and Vertices:

Find center (midpoint of vertices or foci)
Calculate \( a \) from vertices
Calculate \( c \) from foci
Use \( b^2 = a^2 - c^2 \) to find \( b \)
Write equation

📝 Example - From Foci and Vertices:

Write the equation of an ellipse with vertices at \( (\pm 5, 0) \) and foci at \( (\pm 3, 0) \).

Step 1: Center is at origin \( (0, 0) \)
Step 2: \( a = 5 \) (distance from center to vertex)
Step 3: \( c = 3 \) (distance from center to focus)
Step 4: Find \( b \):
\( b^2 = a^2 - c^2 = 25 - 9 = 16 \), so \( b = 4 \)
Step 5: Major axis is horizontal
Equation: \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \)

Converting from General to Standard Form

Method: Completing the Square

General form: \( Ax^2 + By^2 + Cx + Dy + E = 0 \)

Group x-terms and y-terms separately
Factor out coefficients of \( x^2 \) and \( y^2 \) if needed
Complete the square for both x and y
Move constant to right side
Divide by the constant to get 1 on the right side
Simplify to standard form

📝 Example - Completing the Square:

Convert \( 4x^2 + 9y^2 - 8x + 36y + 4 = 0 \) to standard form

Step 1: Group terms:
\( (4x^2 - 8x) + (9y^2 + 36y) = -4 \)

Step 2: Factor out coefficients:
\( 4(x^2 - 2x) + 9(y^2 + 4y) = -4 \)

Step 3: Complete the square:
For x: \( (\frac{-2}{2})^2 = 1 \)
For y: \( (\frac{4}{2})^2 = 4 \)
\( 4(x^2 - 2x + 1) + 9(y^2 + 4y + 4) = -4 + 4(1) + 9(4) \)
\( 4(x - 1)^2 + 9(y + 2)^2 = 36 \)

Step 4: Divide by 36:
\( \frac{(x - 1)^2}{9} + \frac{(y + 2)^2}{4} = 1 \)

Center: \( (1, -2) \), \( a = 3, b = 2 \)

📝 Example 2 - More Practice:

Convert \( 9x^2 + 4y^2 + 54x - 16y + 61 = 0 \)

Group: \( (9x^2 + 54x) + (4y^2 - 16y) = -61 \)
Factor: \( 9(x^2 + 6x) + 4(y^2 - 4y) = -61 \)
Complete square: \( 9(x^2 + 6x + 9) + 4(y^2 - 4y + 4) = -61 + 81 + 16 \)
\( 9(x + 3)^2 + 4(y - 2)^2 = 36 \)
Divide by 36: \( \frac{(x + 3)^2}{4} + \frac{(y - 2)^2}{9} = 1 \)

Center: \( (-3, 2) \), major axis is vertical

Finding Properties from General Form

Direct Method (without converting):

From \( Ax^2 + By^2 + Cx + Dy + E = 0 \):

Center:

\( h = -\frac{C}{2A}, \quad k = -\frac{D}{2B} \)

Note: For other properties, it's easier to convert to standard form first.

⚡ Quick Summary

Property	Formula/Value
Standard Form (Horizontal)	\( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)
Standard Form (Vertical)	\( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \)
Center	\( (h, k) \)
Major Axis Length	\( 2a \)
Minor Axis Length	\( 2b \)
Foci Distance	\( c^2 = a^2 - b^2 \)

\( a \) is always the larger value (semi-major axis)
Major axis is where the larger denominator appears
Vertices are on the major axis, co-vertices on the minor axis
Foci are always on the major axis, inside the ellipse
Use completing the square to convert general to standard form
Sum of distances from any point on ellipse to both foci is constant = \( 2a \)

📚 Important Relationships

Key Equation: