Distance between vertices

**viet** · July 17th 2007, 08:19 AM

Determine the distance D between the vertices of $-9x^{2} + 18 x + 4<br /> y^{2} + 24 y - 9 = 0$
$D =$

**topsquark** · July 17th 2007, 09:12 AM

Originally Posted by viet

Determine the distance D between the vertices of $-9x^{2} + 18 x + 4<br /> y^{2} + 24 y - 9 = 0$
$D =$

This is an hyperbola. First get it into standard form by completing the square on both the x and y terms:
$-9x^{2} + 18 x + 4<br /> y^{2} + 24 y - 9 = 0$

$-(9x^{2} - 18 x) + (4y^{2} + 24y) = 9$

$-9(x^2 - 2x) - 4(y^2 + 6y) = 9$

$-9 ( x^2 - 2x + 1 ) + 9 + 4(y^2 + 6y + 9) - 36 = 9$

$-9(x - 1)^2 + 4(y + 3)^2 = 36$

$-\frac{(x - 1)^2}{4} + \frac{(y + 3)^2}{9} = 1$

$-\frac{(x - 1)^2}{2^2} + \frac{(y + 3)^2}{3^2} = 1$

Can you finish the problem?

-Dan

**viet** · July 17th 2007, 10:40 AM

yeah got it thanks

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