Question 572346
Solve and graph
9x^2-4y^2+18x+32y-91=0
complete the square
9(x^2+2x+1)-4(y^2-8y+16)=91+9-64=36
9(x+1)^2-4(y-4)^2=36
(x+1)^2/4-(y-4)^2/9=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, (h,k) being the (x,y) coordinates of the center.
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For given equation:
Center:(-1,4)
a^2=4
a=2
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b^2=9
b=3
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slope of asymptotes=±b/a=±3/2
Equation of asymptotes:
y=mx+b
y=-3x/2+b
solving for b using (x,y) coordinates of center thru which the asymptote goes.
4=-(3*-1)/2+b
b=5/2
Equation:y=-3x/2+5/2
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y=3x/2+b
solving for b using (x,y) coordinates of center thru which the asymptote goes.
4=(3*-1)/2+b
b=11/2
Equation:y=3x/2+11/2
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see the graph below as a visual check on the above:
y=(2.25(x+1)^2-9)^.5+4
{{{ graph( 300, 300, -10, 10, -10, 10,(2.25(x+1)^2-9)^.5+4,-(2.25(x+1)^2-9)^.5+4,-3x/2+5/2,3x/2+11/2) }}}