1. **State the problem:** We are given the equation $$4x^2 + y^2 - 32x + 4y + 64 = 0$$ and need to analyze it. 2. **Rewrite the equation grouping x and y terms:** $$4x^2 - 32x + y^2 + 4y + 64 = 0$$ 3. **Complete the square for the x terms:** Factor out 4 from the x terms: $$4(x^2 - 8x) + y^2 + 4y + 64 = 0$$ Complete the square inside the parentheses: $$x^2 - 8x = (x - 4)^2 - 16$$ So, $$4[(x - 4)^2 - 16] + y^2 + 4y + 64 = 0$$ 4. **Complete the square for the y terms:** $$y^2 + 4y = (y + 2)^2 - 4$$ 5. **Substitute back and simplify:** $$4(x - 4)^2 - 64 + (y + 2)^2 - 4 + 64 = 0$$ Combine constants: $$4(x - 4)^2 + (y + 2)^2 - 4 = 0$$ Add 4 to both sides: $$4(x - 4)^2 + (y + 2)^2 = 4$$ 6. **Divide both sides by 4 to get standard form:** $$\frac{4(x - 4)^2}{4} + \frac{(y + 2)^2}{4} = \frac{4}{4}$$ Use \cancel to show cancellation: $$\frac{\cancel{4}(x - 4)^2}{\cancel{4}} + \frac{(y + 2)^2}{4} = 1$$ Simplify: $$ (x - 4)^2 + \frac{(y + 2)^2}{4} = 1$$ 7. **Interpretation:** This is the equation of an ellipse centered at $(4, -2)$ with semi-major axis length 2 along the y-axis and semi-minor axis length 1 along the x-axis. **Final answer:** $$\frac{(x - 4)^2}{1} + \frac{(y + 2)^2}{4} = 1$$