|Conic Sections: Hyperbolas
In this lesson you will learn how to write equations of hyperbolas and graphs of hyperbolas will be compared to their equations. Definition
: A hyperbola
is all points found by keeping the difference of the distances from two points (each of which is called a focus
of the hyperbola) constant. The midpoint of the segment
(the transverse axis
) connecting the foci is the center
of the hyperbola. Am hyperbola can be formed by slicing a right circular cone
with a plane
traveling parallel to the vertical axis
of the cone. This effect can be seen in the following video
and screen captures. Part I: Hyperbolas center at the origin.
Example #1: In the first example the constant distance mentioned above will be 6, one focus will be at the point (0, 5) and the other will be at the point (0, -5).The graph of a hyperbola with these foci and center at the origin is shown below. Part II - Hyperbolas translated away from the center
An equation of this hyperbola can be found by using the distance formula. We calculate the distance from the point on the ellipse (x, y) to the two foci, (0, 5) and (0, -5). This total distance is 6 in this example:
Note that 6 is the total distance from vertex to vertex through the center of this hyperbola, shown by the dark red line on the graph. This is called the transverse axis.
After eliminating radicals and simplifying we have Note: a complete derivation of this equation is provided at the bottom of this lesson.
If we let a = 4 and b = 3, this equation can be written as . The important features are:
Example #2: The graph shown below shows the hyperbola .
- b = 3, the distance from the center to the vertices of the hyperbola in the vertical direction (up and down from the center). This is called the transverse axis.
- a = 4, and the two dotted lines in the graph are given by the equations
- is the distance from the center to each focus. Each focus is found on the transverse axis.
- The two dotted lines on the graph are asymptotes because the two branches of the hyperbola approach but never reach these lines. Sketching them first provides a good way to sketch the graph of a hyperbola.
In this case, the transverse axis is horizontal, a = 4, b = 3 and . The hyperbola is not quite the same: it is narrower, and the transverse axis is horizontal rather than vertical. The slopes of the asymptotes are still . The foci are located in the diagram at (-5, 0) and (5, 0), just beyond the vertices (-4, 0) and (4, 0).
The equation of a hyperbola takes the form or where in both cases and locates each focus a distance of c from the center (origin) along the transverse axis. The important features are:
Using this as a model, other equations describing hyperbolas with centers at the origin can be written.
- The transverse axis is in the vertical direction if the y2 term is positive and in the horizontal direction if the x2 term is positive.
- The branches of the hyperbola open up/down if the y2 term is positive and left/right if the x2 term is positive.
- The slopes of the asymptotes are given by
Example #3: If the horizontal distance from the center to the vertices is a = 3, the vertical distance from the center to the vertices is b = 4, and the hyperbola opens left/right, then the equation is Each focus is a distance of from the center. Note that each focus is found horizontally from the center since in this case the transverse axis of the hyperbola is horizontal. A graph from a calculator screen is shown below with the branches of the hyperbola wrapping around each focus. Notice that the calculator does not always join the top and bottom halves of the hyperbolas because of the way it is plotting these points. The dotted lines are the asymptotes which have equations .
Example #4: If then the transverse axis is vertical, a = 6, b = 2, and each focus is a distance of from the center. A graph of this hyperbola is shown below.
The asymptotes are since . One focus is at (0, ) and the other is at (0, ), neither of which shows in the window of this graph since which is outside the vertical window of this graph.
Example #5: Suppose the center is not at the origin (0, 0) but is at some other point such as (2, -1). To graph this hyperbola requires us to remember how graphs are moved horizontally and vertically by a change in the equation. Part III - Summary
Using Example #1 above, we have . This will move the graph in our previous example 2 units right and 1 unit down. Both graphs are shown below.
Note that a = 3, b = 4, , and the slope of the asymptotes is . In this graph the transverse axis is horizontal. Thus each focus is a distance of 5 horizontally from the center. One focus is at (7, -1) and one is at (-3, -1). Note that the graphing calculator does not do a good job of showing the top and bottom halves of the branches of the hyperbola joining at the vertices which are located at (-3, -1) and (5, -1).
Using this as a model, other equations describing hyperbolas with centers at (2, -1) can be written.
- If a = 3 and b = 2, and the transverse axis is horizontal, the equation is
The slope of the asymptotes is and the vertices are located at (-1, -1) and (5, -1).
. The foci are located at .
- If a = 2 and b = 4, and the transverse axis is vertical, the equation is .
The slope of the asymptotes is and the vertices are (2, 3) and (2, -5). . The foci are located at .
To summarize, the equation of a hyperbola is written by using the standard formulas
- the hyperbola opens left and right (transverse axis is horizontal) if the term is positive
- the hyperbola opens up and down (transverse axis is vertical) if the term is positive
- is the distance from the center (h , k) to each focus.
- asymptotes have slopes given by .
- the equations of the asymptotes are given by since the asymptotes contain the center (h, k).
Using these standard equation
forms, answer the following questions about the equation
of a hyperbola.
We calculate the distance from the point on the hyperbola (x, y) to the two foci, (0, c) and (0, -c). The difference of these distances in a constant 2b which is the distance between the vertices found on the transverse axis. In this case the transverse axis is vertical and we use b to describe the distance vertically from the center to a vertex. When we are finished, the y2 term will be positive.
which is the same as
Square both sides:
Expand. Then subtract x2 from both sides:
Subtract y2, 2cy, c2, and 4b2 from both sides:
Divide both sides by 4:
Square both sides:
Subtract b2y2, b2x2, b4, and 2cb2y from both sides:
Since , then and we have
Divide both sides by and we have
If the transverse axis
is horizontal, the difference of the distances from (x, y) to the foci is 2a. The same derivation would give
. It is still true that
. In this case, c is the distance from the center (origin) to the foci, which lie on a horizontal transverse axis.
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