Question 1209669
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.