Hyperbola (H)

A hyperbola is a type of conic section that is formed by the intersection of a right circular conical surface and a plane that cuts through both halves of the cone. It is the set of all points in the plane such that the difference of the distances from two fixed points (foci) is constant.

Standard Equation of Hyperbola

The standard equation of a hyperbola centered at the origin with its transverse axis along the x-axis is given by:

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

where $a$ is the distance from the center to a vertex on the x-axis, and $b$ is the distance from the center to a co-vertex on the y-axis.

For a hyperbola with its transverse axis along the y-axis, the standard equation is:

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

Key Components of a Hyperbola

Center (C): The point at the center of symmetry of the hyperbola.
Vertices (V): The points where the hyperbola intersects its transverse axis.
Foci (F): The two fixed points used to define the hyperbola.
Transverse Axis: The line segment that passes through the foci.
Conjugate Axis: The line segment perpendicular to the transverse axis through the center.
Asymptotes: Lines that the hyperbola approaches but never reaches.

Important Formulas

Distance between foci: $2c$, where $c^2 = a^2 + b^2$.
Length of the transverse axis: $2a$.
Length of the conjugate axis: $2b$.
Equations of asymptotes for horizontal hyperbola: $y = \pm \frac{b}{a}x$.
Equations of asymptotes for vertical hyperbola: $x = \pm \frac{b}{a}y$.

Differences and Important Points

Feature	Ellipse	Hyperbola
Shape	Closed curve	Open curve
Equation	$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$	$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$
Foci	Inside the curve	Outside the curve
Asymptotes	None	Exist and are crossed by the hyperbola
Eccentricity (e)	$0 < e < 1$	$e > 1$
Directrices	Exist and are not crossed by the ellipse	Exist and are not crossed by the hyperbola

Examples

Example 1: Finding the Asymptotes

Given the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$, find the equations of its asymptotes.

Solution:

The hyperbola is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $a^2 = 16$ and $b^2 = 9$. Therefore, $a = 4$ and $b = 3$.

The equations of the asymptotes are $y = \pm \frac{b}{a}x = \pm \frac{3}{4}x$.

Example 2: Finding the Foci

Given the hyperbola $\frac{y^2}{25} - \frac{x^2}{16} = 1$, find the coordinates of its foci.

Solution:

The hyperbola is of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, where $a^2 = 25$ and $b^2 = 16$. Therefore, $a = 5$ and $b = 4$.

We find $c$ using $c^2 = a^2 + b^2 = 25 + 16 = 41$, so $c = \sqrt{41}$.

The foci are located at $(0, \pm c)$, so the coordinates of the foci are $(0, \pm \sqrt{41})$.

Example 3: Graphing a Hyperbola

Graph the hyperbola given by $\frac{x^2}{9} - \frac{y^2}{4} = 1$.

Solution:

Identify $a^2 = 9$ and $b^2 = 4$, which gives $a = 3$ and $b = 2$.
Calculate $c$ using $c^2 = a^2 + b^2 = 9 + 4 = 13$, so $c = \sqrt{13}$.
Plot the center at the origin (0,0).
Plot the vertices at $(\pm a, 0)$, which are $(\pm 3, 0)$.
Plot the foci at $(\pm c, 0)$, which are $(\pm \sqrt{13}, 0)$.
Draw the asymptotes with equations $y = \pm \frac{b}{a}x = \pm \frac{2}{3}x$.
Sketch the hyperbola opening to the left and right, approaching but never touching the asymptotes.

By understanding these components and practicing with examples, students can gain a solid grasp of hyperbolas and be well-prepared for exams.