Soo the equation is 25y^2-9x^2-200y-36x+139=0
and for the center is got (2,4) and the vertices I got (2,7) and (2,1) But i cant figure out how to find the slopes of the asymptotes and it wants the answer is reduced fraction please explain
To find the slopes of the asymptotes of a hyperbola, you need to rewrite the equation of the hyperbola in a specific form called the standard form. The standard form of a hyperbola equation is:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
Where (h, k) represents the coordinates of the center of the hyperbola.
In your case, the equation of the hyperbola is given as 25y^2 - 9x^2 - 200y - 36x + 139 = 0, with center (2, 4). To start, we need to complete the square for both the x and y terms.
For the x terms:
25y^2 - 9x^2 - 200y - 36x + 139 = 0
Rearrange the terms:
-9x^2 - 36x + 25y^2 - 200y + 139 = 0
Group the x and y terms separately:
(-9x^2 - 36x) + (25y^2 - 200y) + 139 = 0
Now complete the square for both x and y terms:
For the x terms:
-9(x^2 + 4x) = -9(x^2 + 4x + 4 - 4) = -9((x + 2)^2 - 4)
For the y terms:
25(y^2 - 8y) = 25(y^2 - 8y + 16 - 16) = 25((y - 4)^2 - 16)
Substitute these results back into the original equation:
-9((x + 2)^2 - 4) + 25((y - 4)^2 - 16) + 139 = 0
-9(x + 2)^2 + 36 + 25(y - 4)^2 - 400 + 139 = 0
-9(x + 2)^2 + 25(y - 4)^2 - 225 = 0
Now, divide the equation by -225 to simplify:
9(x + 2)^2 / 225 - 25(y - 4)^2 / 225 = 1
Comparing this form to the standard form, we have:
a^2 = 225 and b^2 = 9
The square root of a^2 gives us the distance between the center and vertices, and the square root of b^2 gives us the distance between the center and the foci.
So, the distance between the center and vertices is √225 = 15.
The distance between the center and the foci is √9 = 3.
Now, we can find the slopes of the asymptotes. The slopes of the asymptotes for this hyperbola are determined by the ratio of b/a.
Slope of the asymptotes (m) = ± b / a
So, m = ± √(b^2 / a^2) = ± √(9 / 225) = ± 1/5.
Therefore, the slopes of the asymptotes of this hyperbola are ± 1/5 or 1/5 and -1/5.
Note: The slopes of the asymptotes always have a positive and negative value.