1. **Problem:** Determine the type and canonical form of the conic given by the equation $$4x^2 + 2y^2 - 16x + 4y + 10 = 0$$. 2. **Step 1: Identify the conic type.** The general quadratic form is $$Ax^2 + By^2 + Cx + Dy + E = 0$$. Here, $$A=4$$, $$B=2$$, $$C=-16$$, $$D=4$$, $$E=10$$. Since $$A$$ and $$B$$ are both positive and nonzero, this is an ellipse or circle. 3. **Step 2: Complete the square for $$x$$ and $$y$$ terms.** Group $$x$$ and $$y$$ terms: $$4x^2 - 16x + 2y^2 + 4y = -10$$ Divide to simplify: $$4(x^2 - 4x) + 2(y^2 + 2y) = -10$$ Complete the square inside each parenthesis: - For $$x^2 - 4x$$, half of $$-4$$ is $$-2$$, square is $$4$$. - For $$y^2 + 2y$$, half of $$2$$ is $$1$$, square is $$1$$. Add and subtract inside the equation: $$4(x^2 - 4x + 4 - 4) + 2(y^2 + 2y + 1 - 1) = -10$$ Rewrite: $$4((x - 2)^2 - 4) + 2((y + 1)^2 - 1) = -10$$ Expand: $$4(x - 2)^2 - 16 + 2(y + 1)^2 - 2 = -10$$ Combine constants: $$4(x - 2)^2 + 2(y + 1)^2 - 18 = -10$$ Add 18 to both sides: $$4(x - 2)^2 + 2(y + 1)^2 = 8$$ 4. **Step 3: Divide both sides by 8 to get standard form:** $$\frac{(x - 2)^2}{2} + \frac{(y + 1)^2}{4} = 1$$ 5. **Step 4: Conclusion:** This is an ellipse centered at $$(2, -1)$$ with semi-major axis $$2$$ (along $$y$$) and semi-minor axis $$\sqrt{2}$$ (along $$x$$). **Final canonical form:** $$\frac{(x - 2)^2}{2} + \frac{(y + 1)^2}{4} = 1$$