SOLUTION: Find the standard equation,center,foci,asymptotes of 3x^2-2y^2-42x-10y=-64

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Question 1043408: Find the standard equation,center,foci,asymptotes of 3x^2-2y^2-42x-10y=-64
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Find the standard equation,center,foci,asymptotes of
3x%5E2-2y%5E2-42x-10y=-64+...group alike terms
%283x%5E2-42x%29-%282y%5E2%2B10y%29=-64..........complete squares
3%28x%5E2-14x%2B_%29-2%28y%5E2%2B5y%2B_%29=-64+
3%28x%5E2-14x%2B7%5E2%29-3%2A7%5E2-2%28y%2B5%2F2%29%5E2-2%2A%2825%2F4%29=-64+
3%28x-7%29%5E2-3%2A49-2%28y%2B5%2F2%29%5E2-%2825%2F2%29=-64+
3%28x-7%29%5E2-147-2%28y%2B5%2F2%29%5E2-%2825%2F2%29=-64
3%28x-7%29%5E2-2%28y%2B5%2F2%29%5E2=-64+%2B147%2B25%2F2......both sides multiply by 2
6%28x-7%29%5E2-4%28y%2B5%2F2%29%5E2=-128+%2B296%2B25
6%28x-7%29%5E2-4%28y%2B5%2F2%29%5E2=193.....both sides divide by 193
6%28x-7%29%5E2%2F193-4%28y%2B5%2F2%29%5E2%2F193=193%2F193
%286%2F193%29%28x-7%29%5E2-%284%2F193%29%28y%2B5%2F2%29%5E2=1+
%28x-7%29%5E2%2F%28193%2F6%29-%28y%2B5%2F2%29%5E2%2F%28193%2F4%29=1+->hyperbola, the standard equation
h=7
k=+-5%2F2
a+=+sqrt%28193%2F6%29 and b+=sqrt%28193%2F4%29
c%5E2+=+a%5E2+%2B+b%5E2
c%5E2+=%28sqrt+%28193%2F6%29%29%5E2+%2B+%28sqrt%28193%2F4%29%29%5E2
c%5E2+=193%2F6+%2B+193%2F4
c%5E2+=965%2F12
c=sqrt%28965%2F12%29
so, the center is at (h,k)= (7,-5%2F2)
foci: since c+=+sqrt%28965%2F12%29 , the foci will be sqrt%28965%2F12%29 units to either side of the center, and it must be at
(7-sqrt%28965%2F12%29, -5%2F2) and (7%2Bsqrt%28965%2F12%29, -5%2F2)
or (-1.97,-2.5) and (15.96, -2.5)
Since the vertices are a+=sqrt%28+193%2F6%29 units to either side of the center (h,k)= (7,-5%2F2), then they are at
(7-sqrt%28193%2F6%29, -5%2F2) and at (7%2Bsqrt%28193%2F6%29, -5%2F2)
or (1.33, -2.5) and at (12.67, -2.5)
Since the a%5E2 went with the x part of the equation, then a=sqrt%28193%2F6%29 is in the denominator of the slopes of the asymptotes, giving me m+=b%2Fa=+sqrt%28193%2F4%29%2F+sqrt%28193%2F6%29=sqrt%286%2F4%29=sqrt%283%2F2%29; so, m=sqrt%283%2F2%29or m=-sqrt%283%2F2%29. Keeping in mind that the asymptotes go through the center of the hyperbola, the asymptes are then given by the straight-line equations

The asymptotes are y=%28b%2Fa%29%28x-h%29%2Bk, and they are:
y=+sqrt%28+3%2F2%29%28x+-7+%29+-5%2F2+
and
y=++-sqrt%283%2F2%29%28x+-7+%29+-5%2F2+