Ok, the problem is to convert this to standard form...
9x^2-4y^2-18x+16y+11=0, but when I work it out, I get 9(x^2-2x=1)-4(y^2-4Y+4)=4...
This is not going to work out!
What is the mistake I am making or what am I not seeing?
Thanks!
Stacey
Ok, the problem is to convert this to standard form...
9x^2-4y^2-18x+16y+11=0, but when I work it out, I get 9(x^2-2x=1)-4(y^2-4Y+4)=4...
This is not going to work out!
What is the mistake I am making or what am I not seeing?
Thanks!
Stacey
$\displaystyle 9x^2-4y^2-18x+16y+11=0$
$\displaystyle 9x^2-4y^2-18x+16y = -11$
$\displaystyle 9(x^2 - 2x) - 4(y^2 - 4y) = -11$
$\displaystyle 9(x^2 - 2x + 1) - 4(y^2 - 4y + 4) = -11 + 9 - 16$
$\displaystyle 4(y-2)^2 - 9(x-1)^2 = 18$
$\displaystyle \frac{2(y-2)^2}{9} - \frac{(x-1)^2}{2} = 1$