Question 688869: How do i find the Center, Vertices, Foci and the Equation of the asymptotes of the following equation
(((x-2)^2)/25)-(((y-1)^2)/9) = 1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! How do i find the Center, Vertices, Foci and the Equation of the asymptotes of the following equation
(((x-2)^2)/25)-(((y-1)^2)/9) = 1
This is an equation of a hyperbola with horizontal transverse axis. (x-term listed first)
Its standard form of equation:  , (h,k)=(x,y) coordinates of center
For given equation: (x-2)^2)/25)-(y-1)^2)/9)=1
center: (2,1)
a^2=25
a=√25=5
vertices:(2±a,1)=(2±5,1)=(-3,1) and (7,1)
b^2=9
b=√9=3
c^2=a^2+b^2=25+9=34
c=√34≈5.8
foci:(2±c,1)=(2±5.8,1)=(-3.8,1) and (7.8,1)
asymptotes: (straight line equations that go thru center, y=mx+b, m=slope, b=y-intercept)
slopes of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±3/5
Equation of asymptote with negative slope:
y=-3x/5+b
solve for b using coordinates of center(2,1)
1=-3*2/5+b
b=1+6/5=11/5
equation: y=-3x/5+11/5
..
Equation of asymptote with positive slope:
y=3x/5+b
solve for b using coordinates of center(2,1)
1=3*2/5+b
b=1-6/5=-1/5
equation: y=-3x/5-1/5
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