📌 What is a Hyperbola?

A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two fixed points (called foci) to any point on the hyperbola is constant. A hyperbola has two separate branches that open away from each other.

Horizontal Transverse Axis:

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

Vertical Transverse Axis:

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

Key Rule - Identifying Orientation:

From Standard Form:

The center is \( (h, k) \), which can be read directly from the equation.

Center = \( (h, k) \)

📝 Examples - Finding Center:

Example 1: \( \frac{(x - 2)^2}{9} - \frac{(y + 3)^2}{16} = 1 \)

Center: \( (2, -3) \)

Example 2: \( \frac{(y + 1)^2}{25} - \frac{(x - 4)^2}{9} = 1 \)

Center: \( (4, -1) \)

Steps to Find Vertices:

1. Identify which term is positive (determines transverse axis)
2. Find \( a \) from the denominator of the positive term: \( a = \sqrt{a^2} \)
3. Move \( a \) units from the center along the transverse axis

📝 Examples - Finding Vertices:

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

\( x \)-term is positive → Horizontal transverse axis
Center: \( (1, -2) \)
\( a^2 = 16 \), so \( a = 4 \)
Vertices: \( (1 + 4, -2) = (5, -2) \) and \( (1 - 4, -2) = (-3, -2) \)

Example 2: \( \frac{(y - 3)^2}{25} - \frac{(x + 1)^2}{4} = 1 \)

\( y \)-term is positive → Vertical transverse axis
Center: \( (-1, 3) \)
\( a^2 = 25 \), so \( a = 5 \)
Vertices: \( (-1, 3 + 5) = (-1, 8) \) and \( (-1, 3 - 5) = (-1, -2) \)

Formulas:

📝 Example - Axis Lengths:

Find the axis lengths for \( \frac{(x + 2)^2}{36} - \frac{(y - 1)^2}{16} = 1 \)

\( a^2 = 36 \), so \( a = 6 \)
\( b^2 = 16 \), so \( b = 4 \)
Transverse axis length: \( 2a = 2(6) = 12 \)
Conjugate axis length: \( 2b = 2(4) = 8 \)

Asymptote Formulas:

📝 Examples - Asymptotes:

Example 1: \( \frac{(x - 3)^2}{16} - \frac{(y + 1)^2}{9} = 1 \)

Horizontal transverse axis
Center: \( (3, -1) \), \( a = 4, b = 3 \)
\( y - (-1) = \pm\frac{3}{4}(x - 3) \)
Asymptotes:
\( y = \frac{3}{4}(x - 3) - 1 \) and \( y = -\frac{3}{4}(x - 3) - 1 \)
or: \( y = \frac{3}{4}x - \frac{13}{4} \) and \( y = -\frac{3}{4}x + \frac{5}{4} \)

Example 2: \( \frac{y^2}{25} - \frac{x^2}{16} = 1 \)

Vertical transverse axis
Center: \( (0, 0) \), \( a = 5, b = 4 \)
Asymptotes: \( y = \pm\frac{5}{4}x \)

The Relationship Formula:

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

📝 Examples - Finding Foci:

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

Horizontal transverse axis
Center: \( (2, -1) \)
\( a^2 = 9 \), so \( a = 3 \); \( b^2 = 16 \), so \( b = 4 \)
\( c^2 = 9 + 16 = 25 \), so \( c = 5 \)
Foci: \( (2 + 5, -1) = (7, -1) \) and \( (2 - 5, -1) = (-3, -1) \)

Example 2: \( \frac{(y - 4)^2}{25} - \frac{(x + 3)^2}{9} = 1 \)

Vertical transverse axis
Center: \( (-3, 4) \)
\( a = 5, b = 3 \)
\( c^2 = 25 + 9 = 34 \), so \( c = \sqrt{34} \)
Foci: \( (-3, 4 + \sqrt{34}) \) and \( (-3, 4 - \sqrt{34}) \)

Steps:

1. Identify the center \( (h, k) \) from the graph
2. Determine orientation (which way the hyperbola opens)
3. Find \( a \) by measuring from center to a vertex
4. Find \( b \) by measuring from center to a co-vertex OR use asymptote slope
5. Write equation with correct form based on orientation

📝 Example - From Graph:

A hyperbola has center at \( (1, -2) \), opens horizontally, vertices at \( (4, -2) \) and \( (-2, -2) \), and asymptotes with slopes \( \pm\frac{2}{3} \). Write the equation.

Step 1: Center: \( (1, -2) \)
Step 2: Opens horizontally
Step 3: Distance from center to vertex:
\( a = |4 - 1| = 3 \)
Step 4: From asymptote slope \( \frac{b}{a} = \frac{2}{3} \):
\( \frac{b}{3} = \frac{2}{3} \), so \( b = 2 \)
Equation: \( \frac{(x - 1)^2}{9} - \frac{(y + 2)^2}{4} = 1 \)

Given Vertices and Foci:

1. Find center (midpoint of vertices or foci)
2. Calculate \( a \) (distance from center to vertex)
3. Calculate \( c \) (distance from center to focus)
4. Use \( b^2 = c^2 - a^2 \) to find \( b \)
5. Determine orientation and write equation

📝 Example - From Properties:

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

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

Method: Completing the Square

1. Group x-terms and y-terms separately
2. Factor out coefficients if needed
3. Complete the square for both variables
4. Move constant to right side
5. Divide to get 1 on the right side
6. Write in standard form

📝 Example - Completing the Square:

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

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

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

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

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

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

Quick Method for Center:

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

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

Note: For other properties, convert to standard form first using completing the square.

⚡ Quick Summary

• The POSITIVE term determines transverse axis direction
• For hyperbolas: \( c^2 = a^2 + b^2 \) (ADD, not subtract!)
• Foci are always OUTSIDE the vertices (\( c > a \))
• Asymptotes pass through center with slopes \( \pm\frac{b}{a} \) or \( \pm\frac{a}{b} \)
• Hyperbola has TWO separate branches
• Unlike ellipses, \( a \) is NOT necessarily larger than \( b \)

📚 Key Differences: Hyperbola vs Ellipse