1. **State the problem:** We need to analyze and rewrite the conic section given by the equation $$-16x^2 + 25y^2 - 32x - 150y - 191 = 0$$. 2. **Rewrite the equation grouping $x$ and $y$ terms:** $$-16x^2 - 32x + 25y^2 - 150y = 191$$ 3. **Complete the square for $x$ and $y$ terms:** - For $x$: factor out $-16$ from $x$ terms: $$-16(x^2 + 2x)$$ Complete the square inside the parentheses: $$x^2 + 2x + 1 - 1 = (x+1)^2 - 1$$ So, $$-16((x+1)^2 - 1) = -16(x+1)^2 + 16$$ - For $y$: factor out $25$ from $y$ terms: $$25(y^2 - 6y)$$ Complete the square inside the parentheses: $$y^2 - 6y + 9 - 9 = (y-3)^2 - 9$$ So, $$25((y-3)^2 - 9) = 25(y-3)^2 - 225$$ 4. **Substitute back and simplify:** $$-16(x+1)^2 + 16 + 25(y-3)^2 - 225 = 191$$ Combine constants: $$16 - 225 = -209$$ So, $$-16(x+1)^2 + 25(y-3)^2 - 209 = 191$$ Add $209$ to both sides: $$-16(x+1)^2 + 25(y-3)^2 = 400$$ 5. **Divide both sides by 400 to get standard form:** $$\frac{-16(x+1)^2}{400} + \frac{25(y-3)^2}{400} = 1$$ Simplify fractions: $$-\frac{(x+1)^2}{25} + \frac{(y-3)^2}{16} = 1$$ 6. **Rewrite to standard hyperbola form:** Multiply both sides by $-1$: $$\frac{(x+1)^2}{25} - \frac{(y-3)^2}{16} = -1$$ This matches the form of a hyperbola centered at $(-1,3)$ with transverse axis along the $x$-axis. **Final answer:** $$\frac{(x+1)^2}{25} - \frac{(y-3)^2}{16} = -1$$