Hyperbola Analysis

Given the foci at the points (±√29, 0), we can derive the following parameters for the hyperbola:

1. Foci and Eccentricity:

   • The foci can be expressed as (±ae, 0), where ae = √29.
   • Thus, e (eccentricity) can be calculated as: $$e = \frac{\sqrt{29}}{5}$$

2. Calculation of b:

   • Using the relationship $b^2 = a^2(e^2 - 1)$:
   $$b^2 = 25 \left( \frac{29}{25} - 1 \right) = 4$$
   • Therefore, we find:
   $$b = 2$$

3. Equation of the Hyperbola: The standard form of the hyperbola is given by:

   $$\frac{x^2}{25} - \frac{y^2}{4} = 1$$

   The length of the transverse axis is calculated as:

   $$2a = 10$$

4. Finding the Equation: The equation of the hyperbola is provided as:

   $$9x^2 - 16y^2 - 18x + 32y - 151 = 0$$

   Rearranging the equation, we have:

   $$9(x^2 - 2x) - 16(y^2 - 2y) - 151 = 0$$

   Completing the square for both variables:

   $$9\left((x - 1)^2 - 1\right) - 16\left((y - 1)^2 - 1\right) = 151$$

   This simplifies to:

   $$9(x - 1)^2 - 16(y - 1)^2 = 144$$

5. Final Form: Dividing by 144, we arrive at:

   $$\frac{(x - 1)^2}{16} - \frac{(y - 1)^2}{9} = 1$$

6. Center, Eccentricity, and Foci: From this form, we can easily identify:

   • Center: (1, 1)
   • Eccentricity: $\frac{\sqrt{29}}{5}$
   • Foci: (1 ± √29, 1)