Question 515834: Complete the square to determine whether the equation
represents an ellipse, a parabola, a hyperbola, or a degenerate
conic. If the graph is an ellipse, find the center, foci, vertices,
and lengths of the major and minor axes. If it is a parabola, find
the vertex, focus, and directrix. If it is a hyperbola, find the center,
foci, vertices, and asymptotes. Then sketch the graph of the
equation. If the equation has no graph, explain why.
x^2-y^2 = 10(x-y)+1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! It is a hyperbola, find the center,
foci, vertices, and asymptotes.
..
x^2-y^2 = 10(x-y)+1
x^2-y^2 = 10x-10y+1
x^2-10x-y^2+10y=1
complete the square
(x^2-10x+25)-(y^2-10y+25)=1+25-25=1
(x-5)^2-(y-5)^2=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, with (h,k) being the (x,y) coordinates of the center:
For given equation:
center: (5, 5)
a^2=1
a=1
b^2=1
b=1
c^2=a^2+b^2=1+1=2
c=√2
..
foci: (5±c,5)=(5±√2,5)=(5+√2,5) and (5-√2,5)
vertices: (5±a,5)=(5±1,5)=(6,5) and (4,5)
slope:± b/a=±1/1=±1
equation of asymptotes: y=mx+b=±(1)x+b
equation for slope=-1
y=-x+b
solving for b using (x,y) coordinates of center thru which asymptotes pass
5=-5+b
b=10
equation: y=-x+10
..
equation for slope=1
y=x+b
solving for b using (x,y) coordinates of center thru which asymptotes pass
5=5+b
b=
equation: y=x
..
see graph below as a visual check on the answers above:
y=±((x-5)^2-1)^.5+5
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