1. **Problem:** Consider the equation $$2x^2 + 4y^2 + 8x - 16y + F = 0$$. Find all values of $$F$$ such that the graph of the equation a. is an ellipse b. is a point c. consists of no points at all 2. **Rewrite the equation grouping $$x$$ and $$y$$ terms:** $$2x^2 + 8x + 4y^2 - 16y + F = 0$$ 3. **Divide entire equation by 2 to simplify:** $$x^2 + 4x + 2y^2 - 8y + \frac{F}{2} = 0$$ 4. **Complete the square for $$x$$ and $$y$$ terms:** - For $$x^2 + 4x$$, add and subtract $$4$$ (since $$(\frac{4}{2})^2 = 4$$): $$x^2 + 4x + 4 - 4$$ - For $$2y^2 - 8y$$, factor out 2 first: $$2(y^2 - 4y)$$ Complete the square inside parentheses by adding and subtracting $$4$$ (since $$(\frac{4}{2})^2 = 4$$): $$2(y^2 - 4y + 4 - 4) = 2((y - 2)^2 - 4) = 2(y - 2)^2 - 8$$ 5. **Rewrite the equation with completed squares:** $$ (x + 2)^2 - 4 + 2(y - 2)^2 - 8 + \frac{F}{2} = 0$$ 6. **Combine constants:** $$ (x + 2)^2 + 2(y - 2)^2 - 12 + \frac{F}{2} = 0$$ 7. **Isolate the squared terms:** $$ (x + 2)^2 + 2(y - 2)^2 = 12 - \frac{F}{2}$$ 8. **For the graph to be an ellipse, the right side must be positive:** $$12 - \frac{F}{2} > 0 \implies 12 > \frac{F}{2} \implies F < 24$$ 9. **For the graph to be a single point, the right side equals zero:** $$12 - \frac{F}{2} = 0 \implies F = 24$$ 10. **For the graph to have no points, the right side is negative:** $$12 - \frac{F}{2} < 0 \implies F > 24$$ **Final answers:** - a. Ellipse if $$F < 24$$ - b. Point if $$F = 24$$ - c. No points if $$F > 24$$