What is the transverse axis of the following equation?

25x^2+250x-36y^2-576y=2579

25(x^2 + 10x) - 36(y^2 + 16y) = 2579

Complete the squares:

25(x^2 + 10x + 25) - 25*25 - 36(y^2 + 16y + 64) + 36*64 = 2579

25(x+5)^2 - 36(y+8)^2 = 900

How convenient . . .

(x+5)^2/36 - (y+8)^2/25 = 1

Looks like an hyperbola with center at (-5,-8) and horizontal transverse axis

y = -8.

Thank you steve

To determine the transverse axis of this equation, we first need to put it in standard form for a hyperbola. The standard form for a hyperbola with a horizontal transverse axis is:

((x - h)^2 / a^2) - ((y - k)^2 / b^2) = 1

Where:
- (h, k) represents the center of the hyperbola,
- a represents the distance from the center to the vertices along the transverse axis, and
- b represents the distance from the center to the vertices along the conjugate axis.

Let's rearrange the given equation to put it in the standard form:

25x^2 + 250x - 36y^2 - 576y = 2579

First, we can complete the square for the x terms. To do this, we need to divide the equation by 25:

x^2 + 10x - (36y^2 + 576y - 103.16) = 0

Next, we need to complete the square for the y terms. We can do this by dividing the equation by -36:

x^2 + 10x - (-0.9722y^2 + 16y - 28.86) = 0

Now, let's focus on the quadratic expression inside the parentheses. To complete the square, we need to add and subtract the square of half the coefficient of y:

x^2 + 10x - (-0.9722(y^2 - 16y + 64) - 28.86 + 0.9722 * 64) = 0

Simplifying further:

x^2 + 10x + 0.9722(y - 8)^2 - 28.86 + 62.208 = 0

x^2 + 10x + 0.9722(y - 8)^2 + 33.348 = 0

Lastly, we can rearrange the equation to match the standard form:

(x + 5)^2 - 33.348 = -0.9722(y - 8)^2

Now, we can see that the center of the hyperbola is at (-5, 8), and since the equation has a negative term for the y part, we can conclude that the transverse axis is vertical. Therefore, the transverse axis of the given equation is a vertical line passing through (-5, 8).