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Question 581309: Y^2-6y-4x^2-8x=95
Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola. Then graph the hyperbola.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Y^2-6y-4x^2-8x=95
Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola. Then graph the hyperbola.
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Y^2-6y-4x^2-8x=95
complete the square
(y^2-6y+9)-4(x^2+2x+1)=95+9-4
(y-3)^2-4(x+1)^2=100
(y-3)^2/100-4(x+1)^2/25=1
This is an equation of a hyperbola with vertical transverse axis of the standard form:
(y-k)^2/a^2-(x-h)^2/b^2=1, (h,k) being the (x,y) coordinates of the center.
For given hyperbola:
center: (-1,3)
a^2=100
a=10
vertices=(-1,3±a)=(-1,3±10)=(-1,-7) and (-1,13)
..
b^2=25
b=5
..
c^2=a^2+b^2=100+25=125
c=√125≈11.18
Foci=(-1,3±c)=(-1,3±√125)=(-1,3±11.18)= (-1,-8.18) and (-1,14.18)
..
Asymptotes:
Asymptotes are straight lines which go thru the center and have slopes=±a/b
slopes=10/5=2
Standard form of equation for a straight line: y=mx+b, m=slope, b=y-intercept
y=2x+b
solving for b using coordinates of center
3=2*-1+b
b=5
equation: y=2x+5 (asymptote with slope>0)
y=-2x+b
3=-2*-1+b
b=1
equation: y=-2x+1(asymptote with slope<0)
see graph below:
y=±(100+4(x+1)^2)^.5+3
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