SOLUTION: Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, a degenerate conic, or results in no solution. x2 − y2 = 8(x − y)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, a degenerate conic, or results in no solution. x2 − y2 = 8(x − y)      Log On


   



Question 959131: Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, a degenerate conic, or results in no solution.
x2 − y2 = 8(x − y) + 1
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 focus, vertex, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. (Enter your answers for asymptotes as a comma-separated list of equations. If an answer does not exist, enter DNE.)
Sketch the graph of the equation.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+-y%5E2+=+8%28x-y%29+%2B+1
x%5E2+-y%5E2+=+8x-8y+%2B+1
x%5E2-8x+-y%5E2%2B8y+=1
%28x%5E2-8x%2B_%29-_+-%28y%5E2-8y%2B_%29-_+=1 ....since %28a-b%29%5E2=a%5E2-2ab%2Bb%5E2 we see that a=1 and 2ab=8=>2%2A1%2Ab=8=>2b=8=>b=4
%28x%5E2-8x%2B4%29-4-%28y%5E2-8y%2B4%29-4+=1
%28x%5E2-8x%2B2%5E2%29-4-%28y%5E2-8y%2B2%5E2%29-4+=1
%28x-2%29%5E2-%28y-2%29%5E2-8+=1
%28x-2%29%5E2-%28y-2%29%5E2+=1%2B8
%28x-2%29%5E2-%28y-2%29%5E2+=9+
%28x-2%29%5E2%2F9-%28y-2%29%5E2%2F9+=1
so, it is hyperbola %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2+=1+
and we see that
semimajor axis length a=3,
semiminor axis lengthb=3 ,
=> then c=sqrt%283%5E2%2B3%5E2%29=>c=sqrt%282%2A9%29=>c=3sqrt%282%29 or c=-3sqrt%282%29
since h=2,k=2, the center is at (2,2)
foci:
(h-c,+k) | (h%2Bc, k)
((2-3sqrt%282%29, 2) | (2%2B3+sqrt%282%29, 2))
approximately:
(-2.24, 2) | (6.24,2)
vertices:
since the center is at (h, k) = (2, 2) and the vertices are a+=+3 units to either side, then vertices are at
(h-a, k) | (h%2Ba, k)
(-1, 2) | (5, 2)
asymptotes:
y+=+x and +y+=4+-x

Sketch the graph of the equation.