Question

Can someone help me do this with complete the square?

-x^2+y^2-18 x-14 y-132 = 0

Answer 1

do you know what a "perfect square trinomial" is?

Answer 2

i believe i have some of it done actually

-(x2-18x)+81 +(y2-14y)49=132+81+49 ?

Answer 3

yeap

Answer 4

\(\bf -x^2+y^2-18x-14y-132 = 0\\ \quad \\\implies -(x^2+18x)+(y^2-14y)-132 = 0
\\ \quad \\
-(x^2+18x+{\color{red}{ 9}}^2)+(y^2-14y+{\color{red}{ 7}}^2)-132-{\color{red}{ 9}}^2-{\color{red}{ 7}}^2 = 0\)

Answer 5

\(\bf -x^2+y^2-18x-14y-132 = 0
\\ \quad \\
\implies -(x^2+18x)+(y^2-14y)-132 = 0
\\ \quad \\
-(x^2+18x+{\color{red}{ 9}}^2)+(y^2-14y+{\color{red}{ 7}}^2)-132-{\color{red}{ 9}}^2-{\color{red}{ 7}}^2 = 0
\\ \quad \\
-(x+9)^2+(y-7)^2=262\implies (y-7)^2-(x+9)^2=262\)

Answer 6

so after that just 262/262 and distribute it to the other side ?

Answer 7

if you want to get the hyperbola equation, yes

Answer 8

I always get confused on where to do b/2^2 xD

Answer 9

wwould it be better to square root the 262 on the left side? or leave it as is

Answer 10

hmm possible just leave it as it's

Answer 11

Alright thanks!

Answer 12

yw