Determine the distance D between the vertices of $\displaystyle -9x^{2} + 18 x + 4
y^{2} + 24 y - 9 = 0$
$\displaystyle D = $
This is an hyperbola. First get it into standard form by completing the square on both the x and y terms:
$\displaystyle -9x^{2} + 18 x + 4
y^{2} + 24 y - 9 = 0$
$\displaystyle -(9x^{2} - 18 x) + (4y^{2} + 24y) = 9$
$\displaystyle -9(x^2 - 2x) - 4(y^2 + 6y) = 9$
$\displaystyle -9 ( x^2 - 2x + 1 ) + 9 + 4(y^2 + 6y + 9) - 36 = 9$
$\displaystyle -9(x - 1)^2 + 4(y + 3)^2 = 36$
$\displaystyle -\frac{(x - 1)^2}{4} + \frac{(y + 3)^2}{9} = 1$
$\displaystyle -\frac{(x - 1)^2}{2^2} + \frac{(y + 3)^2}{3^2} = 1$
Can you finish the problem?
-Dan