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Question 553690: Write an equation for the hyperbola with vertices (4,0) and (-4,0) and asymptote y=1/4x.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! vertices are (4,0) and (-4,0).
that makes this a horizontally aligned hyperbola.
the center of the hyperbola is midway between the left and right vertex which makes the center equal to (0,0).
the distance from the center to the foci of the hyperbola are called c and are given by the equation c^2 = sqrt(a^2 + b^2).
a is the distance from the vertex to the center which is equal to 4.
the equation for the asymptote of a horizontally aligned hyperbola is y = +/- (b/a) * x
you are given that the equation of the asymptote is y = +/- (1/4)x.
this makes b equal to 1 and a equal to 4.
the formula for a horizontally aligned hyperbola is:
x^2/a^2 - y^2/b^2 = 1 with c = sqrt(a^2 + b^2)
this makes c = sqrt(16+1) = sqrt(17).
this makes the foci at (-sqrt(17),0) and (sqrt(17),0)
so far, the equation of your ellipse is:
x^2/16 - y^2 = 1
to graph this hyperbola, solve for y to get:
y = +/- sqrt((x^2/16)-1)
to graph the asymptotes, graph the equations of:
y = +/- (1/4)x
the graph will look like this:
a more far out view is shown below:
i never remember the formulas and always have to go to the references to get the formula again.
here's a reference i used this time.
http://home.windstream.net/okrebs/page63.html
the foci are not shown on the graph that is mechanically generated.
if you have to show them, then you need to print out the graph and put them there manually.
One of the distinguishing properties of a hyperbola is that the absolute value of the difference of the distance between any point on the hyperbola and the two foci of the hyperbola will be a constant.
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