SOLUTION: The equation of a hyperbola is given by y^2/25 - (x-6)^2/144 = 1.
a. Identify the coordinates of the center of the hyperbola.
b. Find the length of the transverse and conjugate
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Question 454020: The equation of a hyperbola is given by y^2/25 - (x-6)^2/144 = 1.
a. Identify the coordinates of the center of the hyperbola.
b. Find the length of the transverse and conjugate axes.
c. Find the slopes of the asymptotes.
d. Find the coordinates of the foci.
e. Graph the hyperbola. Label the center, midpoints of the associated rectangle, and foci.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
The equation of a hyperbola is given by y^2/25 - (x-6)^2/144 = 1.
a. Identify the coordinates of the center of the hyperbola.
b. Find the length of the transverse and conjugate axes.
c. Find the slopes of the asymptotes.
d. Find the coordinates of the foci.
e. Graph the hyperbola. Label the center, midpoints of the associated rectangle, and foci.
..
y^2/25-(x-6)^2/144=1
This is a hyperbola with vertical transverse axis and center at (6,0)
a^2=25
a=5
coordinates of the vertices: (6,0±5‚
length of transverse axis=2a=10
b^2=144
b=12
length of conjugate axis=2b=24
c^2=a^2+b^2=25+144=169
c=√169=13
coordinates of the foci: (6,0±13)
Slope of Asymptotes:±a/b=±5/12
..
Summary:
a)center (6,0)
b)transverse axis=10, conjugate axis=24
c) Slope of asymptotes ±5/12
d) foci (6,0±13)
e) Graphing hyperbola: I don't have the means to plot the hyperbola in detail, but I believe you have enough information to do this yourself with the aid of the graph below.
..
y=(25+25(x-6)^2/144)^.5
y=5x/12-5/2
y=-5x/12+5/2
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