Question 479586: (x-3)^(2)/(9)-(y+5)^(2)/(16)=1
Identify the curve; find the center,asymptotes, vertices, foci; then sketch the curve
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! (x-3)^(2)/(9)-(y+5)^(2)/(16)=1
Identify the curve; find the center,asymptotes, vertices, foci; then sketch the curve
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(x-3)^(2)/(9)-(y+5)^(2)/(16)=1
This is an equation of a hyperbola with horizontal transverse axis of the standard form:
(x-h)^2/a^2-(y-k)^2/b^2=1
For given equation:
Center: (3,-5)
a^2=9
a=3
length of horizontal transverse axis=2a=6
Vertices:(3±a,-5)=(3±3,-5)=(6,-5) and (0,-5)
..
b^2=16
b=4
length of conjugate axis=2b=8
..
c^2=a^2+b^2=9+16=25
c=√25=5
Foci: (3±c,-5)=(3±5,-5)=(8,-5) and (-2,-5)
..
Asymptotes:
slope, m, =±b/a=±4/3
Equations:
Standard form for straight line: y=mx+b, m=slope, b=y-intercept
y=4x/3+b
solving for b using (x,y) coordinates of center which are on the asymptotes
-5=4*3/3+b
b=-9
Equation: y=4x/3-9
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y=-4x/3+b
-5=-4*3/3+b
b=-1
Equation: y=-4x/3-1
see graph below as a visual check on answers
..
y=±(16+16(x-3)^2/9)^.5-5
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