Can you help me solve this trig problem?

Question

X^2-2Y^2+6X+4Y+5=0 This is a hyperbola.

Answer 1

Glad to hear this. What do you want me to do?
You can always form 2 sums of squares inthe format (X+A)^2 - (Y+B)^2 = C^. Is that what you want?

Answer 2

Complete the square:

(x^2 + 6x) - 2(y^2 - 2y) = -5
(x^2 + 6x + 9) - 2(y^2 - 2y + 1) = -5 + 9 - 2
(x+3)^2 - 2(y-1)^2 = 2

(x+3)^2/2 - (y-1)^2 = 1

The center of the hyperbola is at (-3, 1), a = sqrt(2) and b = 1. So c = sqrt(3) and the vertices are sqrt(2) to the left and right of the center, and the foci are sqrt(3) to the left and right of center.

Answer 3

x^2 - 2y^2 + 6x + 4y + 5 = 0
x^2 + 6x - 2y^2 + 4y + 5 = 0
(x^2 + 6x) - (2y^2 - 4y) + 5 = 0
(x^2 + 6x + 9 - 9) - 2(y^2 - 2y) + 5 = 0
((x + 3)^2 - 9) - 2(y^2 - 2y + 1 - 1) + 5 = 0
(x + 3)^2 - 9 - 2((y - 1)^2 - 1) + 5 = 0
(x + 3)^2 - 2(y - 1)^2 + 2 - 4 = 0
(x + 3)^2 - 2(y - 1)^2 = 2
(((x + 3)^2)/2) - (y - 1)^2 = 1

ANS : (((x + 3)^2)/2) - (y - 1)^2 = 1

of course when you right it, you don't have to use as many ().