Equations of Hyperbolas in Standard Form (Part 2)
(This page is Part 2. Click here for Part 1.)
Be Careful!
The variables $\,a\,,$ $\,b\,$ and $\,c\,$ are used for both ellipses and hyperbolas, but their relationship is different!
For ellipses, $\,c^2 = a^2 \color{red}{-} b^2\,.$
For hyperbolas, $\,c^2 = a^2 \color{red}{+} b^2\,.$
Derivation of the Equation of a Hyperbola: Center at the Origin, Foci on the $y$-axis
The derivation of the equation of a hyperbola with center at the origin and foci on the $y$-axis is nearly identical to the derivation in Part 1.
... the Derivation of the Hyperbola With Center at the Origin and Foci on the Vertical Axis
Summary: Equations of Hyperbolas with Center at the Origin and Foci on the $x$-axis or $y$-axis
In Both Cases:
$0 \lt a \lt c$
With the equations in standard form, the number $\,a^2\,$ is the denominator of the positive term.
The hyperbola constant is $\,2a\,.$
The foci are determined by solving the equation $\,c^2 = a^2 + b^2\,$ for $\,c\,.$
Be Careful!
The variables $\,a\,,$ $\,b\,$ and $\,c\,$ are used for both ellipses and hyperbolas, but their relationship is different!
For ellipses, $\,c^2 = a^2 \color{red}{-} b^2\,.$
For hyperbolas, $\,c^2 = a^2 \color{red}{+} b^2\,.$
Foci on the $x$-axis
Equation of Hyperbola:
$$\cssId{s23}{\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1}$$When the foci are on the $\color{red}{x}$-axis, the $\,\color{red}{x^2}$-term is positive.
Coordinates of foci: $\,(-c,0)\,$ and $\,(c,0)$
Foci on the $y$-axis
Equation of Hyperbola:
$$\cssId{s30}{\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1}$$When the foci are on the $\color{red}{y}$-axis, the $\,\color{red}{y^2}$-term is positive.
Coordinates of foci: $\,(0,-c)\,$ and $\,(0,c)$
Tips
The key to recognizing the equation of a hyperbola with center at the origin and foci on either the $x$-axis or $y$-axis is this:
- It has only $\,x^2\,,$ $\,y^2\,,$ and constant terms.
- When the $\,x^2\,$ and $\,y^2\,$ terms are on the same side of the equation, then they must have different signs.
In both equations, the vertices are easy to find. Since the vertices and foci are both on the same axis (the major axis), this is a good way to minimize memorization and let the equation ‘tell you’ where the foci lie:
- Set $\,y = 0\,$ in the equation $\,\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\,$ to get $\,x = \pm a\,.$ The vertices are on the $x$-axis, hence so are the foci: the major axis is the $x$-axis.
- Set $\,x = 0\,$ in the equation $\,\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\,$ to get $\,y = \pm a\,.$ The vertices are on the $y$-axis, hence so are the foci: the major axis is the $y$-axis.
To find the foci: With the equations in standard form, $\,c^2\,$ is always the sum of the two denominators:
Example: Finding the Equation of a Hyperbola
Find the Equation of the Following Hyperbola:
- Center at the origin
- Major axis is the $y$-axis, with vertices at $\,\pm 5$
- Containing the point $\,(3,5\sqrt{2})$
Also, find the coordinates of the foci.
Solution
Since the major axis is along the $y$-axis, the form of the equation is:
$$ \cssId{s59}{\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1} $$$\,a = 5\,$
Since $\,(3,5\sqrt{2})\,$ is on the graph:
$$ \begin{gather} \cssId{s62}{\frac{(5\sqrt{2})^2}{5^2} - \frac{3^2}{b^2} = 1}\cr\cr \cssId{s63}{2 - \frac{9}{b^2} = 1}\cr\cr \cssId{s64}{\frac{9}{b^2} = 1}\cr\cr \cssId{s65}{b^2 = 9} \end{gather} $$The equation is:
$$ \cssId{s67}{\frac{y^2}{25} - \frac{x^2}{9} = 1} $$Foci:
so $\,c = \sqrt{34}\,.$
Coordinates of foci:
(Note that $\,5 \lt \sqrt{34}\,,$ so the foci are ‘inside’ the hyperbola.)