# Distance between vertices

• July 17th 2007, 09:19 AM
viet
Distance between vertices
Determine the distance D between the vertices of $-9x^{2} + 18 x + 4
y^{2} + 24 y - 9 = 0$

$D =$
• July 17th 2007, 10:12 AM
topsquark
Quote:

Originally Posted by viet
Determine the distance D between the vertices of $-9x^{2} + 18 x + 4
y^{2} + 24 y - 9 = 0$

$D =$

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

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

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

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

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

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

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

Can you finish the problem?

-Dan
• July 17th 2007, 11:40 AM
viet
yeah got it thanks :)