Question 963104: Whats the center focus,asymptotes, and vertexes of these 2 equations
(x-8)^2-(y+6)^2=1
x^2-y^2=4(x-y)+1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Whats the center focus,asymptotes, and vertexes of these 2 equations
(x-8)^2-(y+6)^2=1
This is an equation of a hyperbola with horizontal transverse axis
Its standard form of equation:  , (h,k)=coordinates of center
For given equation:
center: (8, -6)
a^2=1
a=1
b^2=1
b=1
vertices: (8±a,-6)=(8±1,-6)=(7,-6) and (9,-6)
c^2=a^2+b^2=1+1=2
c=√2≈1.4
foci: (8±c,-6)=(8±1.4,-6)=(6.6,-6) and (9.4,-6)
..
asymptotes: straight lines that go thru center
form of equation: y=mx+b, m=slope, b=y-intercept
slope of asymptotes: ±b/a=±1/1=±1
..
equation of asymptote with positive slope:
y=x+b
solve for b using coordinates of center which are on the line
-6=1*8+b
b=-14
equation: y=x-14
..
equation of asymptote with negative slope:
y=-x+b
solve for b using coordinates of center which are on the line
-6=-1*8+b
b=2
equation: y=x+2
..
x^2-y^2=4(x-y)+1
x^2-y^2=4x+4y+1
x^2-4x-y^2-4y=1
complete the square:
(x^2-4x+4)-(y^2-4y+4)=1+4+4
(x-2)^2-(y-2)^2=9
(x-2)^2/9-(y-2)^2/9=1
center: (2,2)
a^2=9
a=3
b^2=9
b=3
vertices: (2±a,2)=(2±3,2)=(1,2) and (5,2)
c^2=a^2+b^2=9+9=18
c=√18≈4.2
foci: (2±c,2)=(2±4.2,2)=(-2.2,2) and (6.2,2)
..
asymptotes: straight lines that go thru center
form of equation: y=mx+b, m=slope, b=y-intercept
slope of asymptotes: ±b/a=±9/9=±1
..
equation of asymptote with positive slope:
y=x+b
solve for b using coordinates of center which are on the line
2=1*2+b
b=0
equation: y=x
..
equation of asymptote with negative slope:
y=-x+b
solve for b using coordinates of center which are on the line
2=-1*2+b
b=4
equation: y=-x+4
|
|
|
