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Question 1209669: Find the asymptote of the hyperbola from part (a) with positive slope.
Enter your answer as an equation of the form .
-4x^2 + y^2 - 2y = -3y^2 + 8x + 9y + 15
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to find the asymptote with a positive slope:
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 and the values of a and b:**
The center of the hyperbola is (-1, 11/8).
a² = 297/64, so a = √(297/64) = (3√33)/8
b² = 297/64, so b = √(297/64) = (3√33)/8
4. **Find the equations of the asymptotes:**
The equations of the asymptotes for a hyperbola in the form (y-k)²/a² - (x-h)²/b² = 1 are given by:
y - k = ±(a/b)(x - h)
In our case:
y - 11/8 = ±((3√33)/8 / (3√33)/8)(x + 1)
y - 11/8 = ±1(x + 1)
y = ±(x+1) + 11/8
5. **Asymptote with positive slope:**
We want the asymptote with the positive slope, so we take the positive sign:
y = x + 1 + 11/8
y = x + 19/8
So, the equation of the asymptote with a positive slope is y = x + 19/8.
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