|
Question 886180: What's the equation of the hyperbola with a focus at (-3,sqrt13), asymptotes intersecting at (-3,1) and one asymptote passing through the point (1,7)?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! as best i can determine, the equation of your hyperbola will be:
(y-1)^2 / 9 - (x+3)^2 / 4 = 1
a^2 = 4
b^2 = 9
c^2 = 13
this makes c = +/- sqrt(13)
as if this isn't confusing enough, some tutorials align the a with the axis line and the b with the line vertical to the axis line and you will see:
a^2 = 9
b^2 = 4
c^2 = 13
the net result is the same, however, with the equation of the hyperbola as shown above.
the center of the hyperbola is the intersection of the asymptotes which is at (-3,1)
the equation of the asymptotes is:
y = 3/2x + 11/2
y = -3/2x - 7/2
the hyperbola will have vertex at (-3,4) and (-3,-2).
this is because the sqrt(9) = 3 and that is the term under the (y-1)^2 term which is equal to the distance from the vertex to the center of the hyperbola.
the width of the hyperbola is equal to the center plus or minus 2 parallel to the x-axis.
the foci of the hyperbola will be at (-3,1+sqrt(13)) and (-3,1-sqrt(13)).
that puts them at about (-3,4.6) and (-3,-2.6)
those focus points are just beyond the vertex points of (-3,4) and (-3,-2)
the eccentricity is equal to c/b or c/a, whichever the case may be.
in this problem it is equal to sqrt(13) / 3 which is equal to 1.2 roughly.
the graph of the hyperbola is shown below:
here's a couple of links that talk about hyperbolas.
http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx
http://www.purplemath.com/modules/hyperbola.htm
|
|
|
| |
