kathryn102
kathryn102 2d ago โ€ข 0 views

Solved examples: Converting ellipse equations from general to standard form

Hey there! ๐Ÿ‘‹ Ellipses can seem tricky, but converting their equations to standard form makes them way easier to understand. Let's break it down with some examples and then test your skills with a quiz! Good luck!๐Ÿ€
๐Ÿงฎ Mathematics

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bryant.teresa34 Dec 30, 2025

๐Ÿ“š Quick Study Guide

  • ๐Ÿ” The general form of an ellipse equation is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, where $B = 0$ and $A$ and $C$ have the same sign.
  • ๐Ÿ“ The standard form of an ellipse equation centered at $(h, k)$ is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ (horizontal major axis) or $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$ (vertical major axis), where $a > b$.
  • ๐Ÿ’ก To convert from general to standard form, complete the square for both $x$ and $y$ terms.
  • โž• Identify the center $(h, k)$ by looking at the values being subtracted from $x$ and $y$ in the standard form.
  • ๐Ÿ“ Determine the lengths of the major and minor axes based on the values of $a$ and $b$. The major axis has length $2a$, and the minor axis has length $2b$.

Practice Quiz

  1. Question 1: Convert the following 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 equation to standard form: $25x^2 + 4y^2 + 150x - 16y + 141 = 0$
    1. $\frac{(x+3)^2}{4} + \frac{(y-2)^2}{25} = 1$
    2. $\frac{(x-3)^2}{4} + \frac{(y+2)^2}{25} = 1$
    3. $\frac{(x+3)^2}{25} + \frac{(y-2)^2}{4} = 1$
    4. $\frac{(x-3)^2}{25} + \frac{(y+2)^2}{4} = 1$
  3. Question 3: Convert the following equation to standard form: $16x^2 + y^2 - 96x + 8y + 144 = 0$
    1. $\frac{(x-3)^2}{1} + \frac{(y+4)^2}{16} = 1$
    2. $\frac{(x+3)^2}{1} + \frac{(y-4)^2}{16} = 1$
    3. $\frac{(x-3)^2}{16} + \frac{(y+4)^2}{1} = 1$
    4. $\frac{(x+3)^2}{16} + \frac{(y-4)^2}{1} = 1$
  4. Question 4: What is the center of the ellipse given by the equation: $9(x-1)^2 + 4(y+2)^2 = 36$?
    1. (1, 2)
    2. (-1, -2)
    3. (1, -2)
    4. (-1, 2)
  5. Question 5: What are the lengths of the major and minor axes of the ellipse given by $\frac{(x+5)^2}{16} + \frac{(y-3)^2}{9} = 1$?
    1. Major axis: 3, Minor axis: 4
    2. Major axis: 8, Minor axis: 6
    3. Major axis: 4, Minor axis: 3
    4. Major axis: 16, Minor axis: 9
  6. Question 6: Convert the following 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. $\frac{(x+3)^2}{4} + \frac{(y-1)^2}{1} = 1$
    4. $\frac{(x-3)^2}{4} + \frac{(y+1)^2}{1} = 1$
  7. Question 7: Which of the following equations represents an ellipse?
    1. $x^2 - y^2 = 1$
    2. $x^2 + y = 1$
    3. $x^2 + y^2 = 1$
    4. $4x^2 + 9y^2 = 36$
Click to see Answers
  1. C
  2. A
  3. A
  4. C
  5. B
  6. A
  7. D

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