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).