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Question 449662: write the equation of the conic section with the given information.
hyperbola with vertices (3,-2) and (-9,-2), and foci (7, -2) and (-13, -2)
Thank you so much for all your help!!!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! write the equation of the conic section with the given information.
hyperbola with vertices (3,-2) and (-9,-2), and foci (7, -2) and (-13, -2)
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Standard form for hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1,
with (h,k) being the (x,y) coordinates of the center.
Standard form for hyperbola with vertical transverse axis: (y-k)^2/a^2-(x-h)^2/b^2=1
with (h,k) being the (x,y) coordinates of the center.
Note that the only difference between these two forms is that the x and y terms are interchanged.
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The best way to start solving conics problems is drawing a sketch of the conic with the given information like the four points of the vertices and foci.
Since the vertices and foci are on the line y=-2, transverse axis of given hyperbola is horizontal, that is, it opens left-right. (First form listed above)
Center of hyperbola is also on the line y=-2, so its y-coordinate=-2.
The x-coordinate of center is the mid point of the end points of the vertices=(3+(-9))/2=6/2=-3.
So, coordinates of center is at (-3,-2)
Length of vertex is the distance between the end points of a vertex=12=2a
a=6
a^2=36
c=distance from center to one of the foci=10
c^2=100
c^2=a^2+b^2
b^2=c^2-a^2=100-36=64
b=8
We now have the information we need to write the equation of given hyperbola in standard form
(x-h)^2/a^2-(y-k)^2/b^2=1
(x+3)^2/36-(y+2)^2/64=1 (ans)
See the graph of given hyperbola below which should help you relate the visual evidence to the algebra above.
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(64((x+3)^2/36-1))^.5-2
(8/6)(x+3)-2 (equation of asymptotes)
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