Determine the distance D between the vertices of $\displaystyle -9x^{2} + 18 x + 4

y^{2} + 24 y - 9 = 0$

$\displaystyle D = $

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- Jul 17th 2007, 08:19 AMvietDistance between vertices
Determine the distance D between the vertices of $\displaystyle -9x^{2} + 18 x + 4

y^{2} + 24 y - 9 = 0$

$\displaystyle D = $ - Jul 17th 2007, 09:12 AMtopsquark
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 - Jul 17th 2007, 10:40 AMviet
yeah got it thanks :)