Understanding the Hyperbola

A hyperbola is a type of conic section that is formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This results in two mirror-image curves, each one being a branch of the hyperbola. The standard form of the equation of a hyperbola centered at the origin with its transverse axis along the x-axis is:

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

where ( a ) is the distance from the center to a vertex and ( b ) is the distance from the center to the co-vertex.

Key Elements of a Hyperbola

Element	Description	Formula (for hyperbola centered at origin)
Center	The point that is equidistant from all vertices and foci.	( (0, 0) )
Vertices	Points where the hyperbola intersects its transverse axis.	( (\pm a, 0) )
Foci (Focus)	Points inside the hyperbola that are used to define the curve.	( (\pm c, 0) ) where ( c^2 = a^2 + b^2 )
Directrix	Lines that are perpendicular to the transverse axis and equidistant from the center.	( x = \pm \frac{a^2}{c} )
Latus Rectum	Line segments perpendicular to the transverse axis that pass through the foci.	Length: ( \frac{2b^2}{a} )
Transverse Axis	The line segment that passes through the foci of the hyperbola.	Length: ( 2a )
Conjugate Axis	The line segment that is perpendicular to the transverse axis and passes through the center.	Length: ( 2b )
Eccentricity (e)	A measure of how much the hyperbola deviates from being circular.	( e = \frac{c}{a} )

Finding the Elements of a Hyperbola

Example 1: Given the Hyperbola Equation

Given the equation of a hyperbola ( \frac{x^2}{16} - \frac{y^2}{9} = 1 ), find the foci, vertices, directrices, and the length of the latus rectum.

Solution:

1. Identify ( a^2 ) and ( b^2 ):

   • ( a^2 = 16 ) so ( a = 4 )
   • ( b^2 = 9 ) so ( b = 3 )

2. Calculate ( c^2 ):

   • ( c^2 = a^2 + b^2 = 16 + 9 = 25 ) so ( c = 5 )

3. Find the vertices:

   • ( (\pm a, 0) = (\pm 4, 0) )

4. Find the foci:

   • ( (\pm c, 0) = (\pm 5, 0) )

5. Find the directrices:

   • ( x = \pm \frac{a^2}{c} = \pm \frac{16}{5} )

6. Calculate the length of the latus rectum:

   • Length: ( \frac{2b^2}{a} = \frac{2 \cdot 9}{4} = \frac{18}{4} = 4.5 )

Example 2: Hyperbola with Vertices and Foci

Given that the vertices of a hyperbola are at ( (0, \pm 6) ) and the foci are at ( (0, \pm 10) ), find the equation of the hyperbola.

Solution:

1. Since the vertices are on the y-axis, the hyperbola is vertical, and its equation will be of the form ( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 ).

2. Identify ( a ) and ( c ):

   • ( a = 6 ) so ( a^2 = 36 )
   • ( c = 10 ) so ( c^2 = 100 )

3. Calculate ( b^2 ):

   • ( b^2 = c^2 - a^2 = 100 - 36 = 64 )

4. Write the equation of the hyperbola:

   • ( \frac{y^2}{36} - \frac{x^2}{64} = 1 )

Example 3: Eccentricity

Given the equation of a hyperbola ( \frac{x^2}{25} - \frac{y^2}{16} = 1 ), find the eccentricity.

Solution:

1. Identify ( a^2 ) and ( b^2 ):

   • ( a^2 = 25 ) so ( a = 5 )
   • ( b^2 = 16 ) so ( b = 4 )

2. Calculate ( c ):

   • ( c^2 = a^2 + b^2 = 25 + 16 = 41 ) so ( c = \sqrt{41} )

3. Find the eccentricity:

   • ( e = \frac{c}{a} = \frac{\sqrt{41}}{5} )

By understanding these elements and how to derive them from the equation of a hyperbola, students can solve a wide range of problems related to this conic section. Practice with different equations and configurations of hyperbolas will help solidify these concepts.