whats the center of the hyperola
25y^2-144x^2+150y-576x-3951=0
Complete the squares or the x and y binomials.
-(12x + 24)^2 +576 + (5y +15)^2 -225 -3951 = 0
-12(x+2)^2 +5(y+3)^2 = 3600
This can be written
-12 x'^2 + 5y'^2 = 3600
where x' = x+2 and y' = y+3
This is the equation of a hyperbola centered at x= -2, y = -3, where x' and y' are zero.
To find the center of the hyperbola represented by the equation 25y^2 - 144x^2 + 150y - 576x - 3951 = 0, we need to rewrite the equation in a standard form.
First, let's group the x and y terms separately:
25y^2 + 150y - 144x^2 - 576x - 3951 = 0
Now, let's complete the square for the x-term by focusing on the x^2 and x terms.
-144x^2 - 576x = -(144x^2 + 576x)
To complete the square for this quadratic expression, we divide the coefficient of x by 2, square it, and add it to both sides of the equation:
-(144x^2 + 576x) = -(144(x^2 + 4x + 4) - 4^2)
Simplifying the expression inside the parentheses:
-(144(x + 2)^2 - 16) = -144(x + 2)^2 + 16
Now let's do the same thing for the y-term by focusing on the y and y^2 terms.
25y^2 + 150y = 25(y^2 + 6y) = 25(y^2 + 6y + 9 - 9) = 25((y + 3)^2 - 9)
Now, let's rewrite the equation in standard form:
25((y + 3)^2 - 9) - 144(x + 2)^2 + 16 - 3951 = 0
Expanding the equation:
25(y + 3)^2 - 225 - 144(x + 2)^2 + 16 - 3951 = 0
Simplifying the equation further:
25(y + 3)^2 - 144(x + 2)^2 - 4160 = 0
We can now rewrite this equation in the standard form:
25(y + 3)^2 - 144(x + 2)^2 = 4160
We can see that this equation is in the form ((y - k)^2 / a^2) - ((x - h)^2 / b^2) = 1, where (h, k) is the center of the hyperbola.
Comparing this with our equation, we can determine the center:
Center: (-2, -3)
Therefore, the center of the hyperbola represented by the equation 25y^2 - 144x^2 + 150y - 576x - 3951 = 0 is (-2, -3).
To find the center of the hyperbola, we need to rewrite the equation in standard form by completing the square for both the x and y terms.
25y^2 - 144x^2 + 150y - 576x - 3951 = 0
Rearranging the terms, we group the x and y terms together:
25y^2 + 150y - 144x^2 - 576x = 3951
Next, let's complete the square for both the x and y terms individually. Starting with the x terms:
25y^2 + 150y - (144x^2 + 576x) = 3951
To complete the square for the x terms, we need to add (144/2)^2 = 1296 to both sides:
25y^2 + 150y - (144x^2 + 576x + 1296) = 3951 + 1296
Simplifying, we have:
25y^2 + 150y - (12x + 18)^2 = 5247
Moving on to the y terms, we need to complete the square for the y terms and add (150/2)^2 = 5625 to both sides:
25(y^2 + 6y + 5625/25^2) - (12x + 18)^2 = 5247 + 5625
Simplifying, we get:
(y + 150/25)^2 - (12x + 18)^2 = 722
Let's simplify a bit more:
(y + 6)^2/25 - (12x + 18)^2 = 722
Dividing both sides by 25, we have:
(y + 6)^2/25 - (12x + 18)^2/25 = 722/25
Now we can write the equation in standard form:
(y + 6)^2/25 - (12x + 18)^2/25 = 722/25
Comparing this to the standard form of a hyperbola equation:
(y - k)^2/a^2 - (x - h)^2/b^2 = 1
We can see that the center of the hyperbola is given by the values (h, k). Therefore, the center of the hyperbola is (-3, -6).