|
Question 1209667: Parts (a)-(c) refer to the graph of
-4x^2 + y^2 - 2y = -3y^2 + 8x + 9y + 15
which is a hyperbola.
Find the center of the hyperbola.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to find the center of the hyperbola:
1. **Simplify and rearrange the equation:**
-4x² + y² - 2y = -3y² + 8x + 9y + 15
-4x² - 8x + 4y² - 11y - 15 = 0
-4(x² + 2x) + 4(y² - (11/4)y) = 15
-4(x² + 2x + 1) + 4 + 4(y² - (11/4)y + (121/64)) - 121/16 = 15
-4(x + 1)² + 4(y - 11/8)² = 15 - 4 + 121/16
-4(x + 1)² + 4(y - 11/8)² = 11 + 121/16
-4(x + 1)² + 4(y - 11/8)² = (176 + 121)/16
-4(x + 1)² + 4(y - 11/8)² = 297/16
2. **Divide to get the standard form:**
-(x + 1)²/(297/64) + (y - 11/8)²/(297/64) = 1
(y - 11/8)²/(297/64) - (x + 1)²/(297/64) = 1
3. **Identify the center:**
The standard form of a hyperbola centered at (h, k) is:
(y - k)²/a² - (x - h)²/b² = 1 (for a vertical hyperbola)
Comparing this with our equation, we can see that:
h = -1
k = 11/8
Therefore, the center of the hyperbola is (-1, 11/8).
|
|
|
| |
