Test questions for converting ellipse general form to standard form (Completing the Square)

📚 Quick Study Guide

• 🔑 The general form of an ellipse is $Ax^2 + Cy^2 + Dx + Ey + F = 0$, where $A$ and $C$ have the same sign but $A \neq C$.
• 📝 To convert from general form to standard form, you'll need to complete the square for both the $x$ and $y$ terms.
• 🔢 The standard form of an ellipse centered at $(h, k)$ is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. If $a > b$, the major axis is horizontal; if $b > a$, the major axis is vertical.
• ➗ Group the $x$ terms, group the $y$ terms, and move the constant to the right side of the equation.
• ➕ Complete the square for both $x$ and $y$. Remember to add the same values to both sides of the equation.
• ➗ Divide both sides of the equation by the constant on the right side to make it equal to 1.
• 🎯 Identify the center $(h, k)$, the values of $a$ and $b$, and whether the major axis is horizontal or vertical.

✍️ Practice Quiz

1. Question 1: Convert the following ellipse equation to standard form: $4x^2 + 9y^2 - 16x + 18y - 11 = 0$
   1. $\frac{(x-2)^2}{9} + \frac{(y+1)^2}{4} = 1$
   2. $\frac{(x+2)^2}{9} + \frac{(y-1)^2}{4} = 1$
   3. $\frac{(x-2)^2}{4} + \frac{(y+1)^2}{9} = 1$
   4. $\frac{(x+2)^2}{4} + \frac{(y-1)^2}{9} = 1$
2. Question 2: Convert the following ellipse equation to standard form: $x^2 + 4y^2 + 6x - 8y + 9 = 0$
   1. $\frac{(x+3)^2}{4} + (y-1)^2 = 1$
   2. $\frac{(x-3)^2}{4} + (y+1)^2 = 1$
   3. $(x+3)^2 + \frac{(y-1)^2}{4} = 1$
   4. $(x-3)^2 + \frac{(y+1)^2}{4} = 1$
3. Question 3: Convert the following ellipse equation to standard form: $9x^2 + 4y^2 + 18x - 16y - 11 = 0$
   1. $\frac{(x+1)^2}{4} + \frac{(y-2)^2}{9} = 1$
   2. $\frac{(x-1)^2}{4} + \frac{(y+2)^2}{9} = 1$
   3. $\frac{(x+1)^2}{9} + \frac{(y-2)^2}{4} = 1$
   4. $\frac{(x-1)^2}{9} + \frac{(y+2)^2}{4} = 1$
4. Question 4: Convert the following ellipse equation to standard form: $25x^2 + 16y^2 + 50x - 64y - 311 = 0$
   1. $\frac{(x+1)^2}{16} + \frac{(y-2)^2}{25} = 1$
   2. $\frac{(x-1)^2}{16} + \frac{(y+2)^2}{25} = 1$
   3. $\frac{(x+1)^2}{25} + \frac{(y-2)^2}{16} = 1$
   4. $\frac{(x-1)^2}{25} + \frac{(y+2)^2}{16} = 1$
5. Question 5: Convert the following ellipse equation to standard form: $4x^2 + y^2 - 8x + 4y - 8 = 0$
   1. $\frac{(x-1)^2}{4} + (y+2)^2 = 1$
   2. $(x-1)^2 + \frac{(y+2)^2}{4} = 1$
   3. $\frac{(x+1)^2}{4} + (y-2)^2 = 1$
   4. $(x+1)^2 + \frac{(y-2)^2}{4} = 1$
6. Question 6: Convert the following ellipse equation to standard form: $16x^2 + 9y^2 - 32x + 36y - 92 = 0$
   1. $\frac{(x-1)^2}{9} + \frac{(y+2)^2}{16} = 1$
   2. $\frac{(x+1)^2}{9} + \frac{(y-2)^2}{16} = 1$
   3. $\frac{(x-1)^2}{16} + \frac{(y+2)^2}{9} = 1$
   4. $\frac{(x+1)^2}{16} + \frac{(y-2)^2}{9} = 1$
7. Question 7: Convert the following ellipse equation to standard form: $x^2 + 9y^2 + 4x - 18y + 4 = 0$
   1. $(x+2)^2 + \frac{(y-1)^2}{9} = 1$
   2. $\frac{(x+2)^2}{9} + (y-1)^2 = 1$
   3. $(x-2)^2 + \frac{(y+1)^2}{9} = 1$
   4. $\frac{(x-2)^2}{9} + (y+1)^2 = 1$