Question 616686: Sketch the hyperbolas. Identify the center, asymptotes, vertices, and foci.
(x-3)^2/9 - (y+5)^2/16 = 1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Sketch the hyperbolas. Identify the center, asymptotes, vertices, and foci.
(x-3)^2/9 - (y+5)^2/16 = 1
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Equation is that of a hyperbola with vertical transverse axis.
Its standard form: (y-k)^2/a^2-(x-h)^2/b^2=1, (h,k)=(x,y) coordinates of the center
For given equation:
center: (0,0)
a^2=16
a=√16=4
vertices: (0, 0±a)=(0,0±4)=(0,-4) and (0,4)
..
b^2=9
b=√9=3
..
c^2=a^2+b^2=16+9=25
c=√25=5
foci:(0, 0±c)=(0,0±5)=(0,-5) and (0,5)
...
slopes of asymptotes with vertical transverse axis=±a/b=±4/3
asymptotes are straight lines which intersect at the center.
equation:y=mx+b, m=slope, b=y-intercept
y-intercept=0, since asymptotes go thru the origin(0,0)
equation of asymptote with negative slope: y=-4x/3
equation of asymptote with positive slope: y=4x/3
..
see graph below:
y=±(16+16x^2/9)^.5
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