Question 959131
{{{x^2 -y^2 = 8(x-y) + 1}}}

{{{x^2 -y^2 = 8x-8y + 1}}}

{{{x^2-8x -y^2+8y =1}}}

{{{(x^2-8x+_)-_ -(y^2-8y+_)-_ =1}}} ....since {{{(a-b)^2=a^2-2ab+b^2}}} we see that {{{a=1}}} and {{{2ab=8}}}=>{{{2*1*b=8}}}=>{{{2b=8}}}=>{{{b=4}}}

{{{(x^2-8x+4)-4-(y^2-8y+4)-4 =1}}}

{{{(x^2-8x+2^2)-4-(y^2-8y+2^2)-4 =1}}}

{{{(x-2)^2-(y-2)^2-8 =1}}}

{{{(x-2)^2-(y-2)^2 =1+8}}}

{{{(x-2)^2-(y-2)^2 =9 }}}

{{{(x-2)^2/9-(y-2)^2/9 =1}}} 

so, it is hyperbola {{{(x-h)^2/a^2-(y-k)^2/b^2 =1 }}}

and we see that 
semimajor axis length {{{a=3}}}, 
semiminor axis length{{{b=3}}} , 

=> then {{{c=sqrt(3^2+3^2)}}}=>{{{c=sqrt(2*9)}}}=>{{{c=3sqrt(2)}}} or {{{c=-3sqrt(2)}}}

since {{{h=2}}},{{{k=2}}}, the center is at ({{{2}}},{{{2}}})

foci: 
({{{h-c}}},{{{ k}}})  |  ({{{h+c}}}, {{{k}}})
(({{{2-3sqrt(2)}}}, {{{2}}})  |  ({{{2+3 sqrt(2)}}}, {{{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+a}}}, {{{k}}})
 ({{{-1}}}, {{{2}}})  |  ({{{5}}}, {{{2}}})

asymptotes: 

{{{y = x}}}  and {{{ y =4 -x}}}


Sketch the graph of the equation.

{{{ graph( 600, 600, -10, 10, -10, 10,x,4-x ,sqrt(((x-2)^2/9-1)9)+2,-sqrt(((x-2)^2/9-1)9)+2) }}}