SOLUTION: x^2-y^2+6x+10y-17=0 A.) Write the equation in standard form. B.) Find the Center, Verticies, Foci, and slope of asymptotes.

Question 455268: x^2-y^2+6x+10y-17=0
A.) Write the equation in standard form.
B.) Find the Center, Verticies, Foci, and slope of asymptotes.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
x^2-y^2+6x+10y-17=0
A.) Write the equation in standard form.
B.) Find the Center, Verticies, Foci, and slope of asymptotes.
..
Standard form of a hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
Standard form of a hyperbola with vertical transverse axis: (y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
The difference between the two forms is the interchange of (x-h) and (y-k)
..
x^2-y^2+6x+10y-17=0
completing the square
(x^2+6x+9)-(y^2-10y+25)=17+9+25=51
(x+3)^2/51-(y-5)^2/51=1
This is a hyperbola with horizontal transverse axis (first form listed above).
center:(-3,5)
a^2=51
a=√51=7.1..
vertices: (-3�√51,5)
..
b^2=51
b=√51
..
c^2=a^2+b^2=51+51=102
c=√(102)=10.1..
..
foci: (-3�√(102),5)
.
asymptotes
slope=�b/a=�√51/√51=�1
Equation of asymptotes
y=mx+b
y=�x+b
solving for b using coordinates of center which are on the asymptotes
5=�3+b
b=8 and 2
Equations:
y=-x+2
y=x+8
See graph below as a visual check on the answers
..
y=((x+3)^2-51)^.5+5

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