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. 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:  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.
Example #2: The graph shown below shows the 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:  The transverse axis is in the vertical direction if the y^{2} term is positive and in the horizontal direction if the x^{2} term is positive.
 The branches of the hyperbola open up/down if the y^{2} term is positive and left/right if the x^{2} term is positive.
 The slopes of the asymptotes are given by
Using this as a model, other equations describing hyperbolas with centers at the origin can be written.
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. Part II  Hyperbolas translated away from the centerExample #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.
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 . Part III  Summary To summarize, the equation of a hyperbola is written by using the standard formulas where  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 and graph of a hyperbola. 


Derivation:
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 y^{2} term will be positive. which is the same as Square both sides: Expand. Then subtract x^{2} from both sides: Subtract y^{2}, 2cy, c^{2}, and 4b^{2} from both sides: Divide both sides by 4: Square both sides: Subtract b^{2}y^{2}, b^{2}x^{2}, b^{4}, and 2cb^{2}y 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.

M Ransom J Anderson


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