SOLUTION: Find the vertices of the hyperbola from part (a). -4x^2 + y^2

SOLUTION: Find the vertices of the hyperbola from part (a). -4x^2 + y^2 - 2y = -3y^2 + 8x + 9y + 15

Question 1209668: Find the vertices of the hyperbola from part (a).

-4x^2 + y^2 - 2y = -3y^2 + 8x + 9y + 15
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to find the vertices of the hyperbola:
1. **Simplify and rearrange the equation (same as in the previous response):**
-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 and the value of a:**
The center of the hyperbola is (-1, 11/8).
a² = 297/64, so a = √(297/64) = (3√33)/8
4. **Find the vertices:**
Since the y² term is positive, the hyperbola opens vertically. The vertices are located a units above and below the center.
Vertices: (-1, 11/8 ± a) = (-1, 11/8 ± (3√33)/8)
Therefore, the vertices of the hyperbola are:
(-1, (11 + 3√33)/8) and (-1, (11 - 3√33)/8)
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