Ellipses

TipLearning Objectives

By the end of this lesson, you’ll be able to:

  • Identify the key features of an ellipse (center, vertices, co-vertices, major/minor axes, foci).
  • Write and interpret ellipses in standard form.
  • Graph ellipses using axis lengths and center coordinates.
  • Connect algebraic parameters to geometric features, a skill often tested on ACT Math.

Key Ideas

An ellipse is the set of points whose total distance to two fixed points (the foci) is constant.

The standard forms:

Horizontal major axis: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]

Vertical major axis: \[ \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \]

Where:

  • \((h,k)\) is the center
  • \(a\) = semi-major axis (longer)
  • \(b\) = semi-minor axis (shorter)
  • Relationship: \[ c^2 = a^2 - b^2 \]
  • Foci:
    • Horizontal: \((h \pm c,\, k)\)
    • Vertical: \((h,\, k \pm c)\)

Ellipse showing the center, major axis, minor axis, and vertices.

Common Problem Types

1. Identifying Orientation

Check whether the larger denominator is under \((x-h)^2\) or \((y-k)^2\).


2. Finding \(a\), \(b\), and \(c\)

Use
\[ c^2 = a^2 - b^2 \]
then write the foci coordinates.


3. Converting from General Form

Complete the square for both \(x\) and \(y\) terms to rewrite in standard form.


4. Graphing from Standard Form

Locate center → major/minor axis lengths → vertices → co-vertices → foci.


5. Missing Parameter Problems

ACT often gives part of the ellipse (like endpoints or foci) and asks you to determine \(a\), \(b\), or the equation.

Strategies

  • Circle or ellipse? If the denominators are equal, it’s a circle.
  • Orientation shortcut: Larger denominator → major axis direction.
  • Check center translations carefully—ACT loves shifting ellipses.
  • Compute \(c\) last to avoid mixing up \(a\) and \(b\).
  • Sketch every time for quick verification.

Worked Examples

Example 1: Identify key features

What are the center and foci of: \[ \frac{(x-2)^2}{25} + \frac{(y+1)^2}{9} = 1? \]

  • Larger denominator is 25 → horizontal major axis
  • Center: \((2, -1)\)
  • \(a^2 = 25 \Rightarrow a = 5\)
  • \(b^2 = 9 \Rightarrow b = 3\)
  • Compute \(c\): \[ \begin{split} c^2 &= a^2 - b^2 \\ &= 25 - 9 \\ &= 16 \end{split} \]
    So \(c = 4\).
  • Foci: \((2 \pm 4, -1)\)\((-2,-1)\) and \((6,-1)\)

Example 2: Write the equation

An ellipse has center \((0,0)\), vertical major axis, \(a=7\), \(b=4\).
Because the major axis is vertical, \(a^2\) goes under the \(y\)-term:

\[ \frac{x^2}{4^2} + \frac{y^2}{7^2} = 1 \]

So the equation is: \[ \frac{x^2}{16} + \frac{y^2}{49} = 1 \]

Example 3: Convert to standard form

Convert
\[ 9x^2 + 4y^2 - 54x + 16y + 61 = 0 \]
to standard form.

Group terms:

\[ 9(x^2 - 6x) + 4(y^2 + 4y) = -61 \]

Complete the square:

\[ \begin{split} x^2 - 6x &= (x-3)^2 - 9 \\ y^2 + 4y &= (y+2)^2 - 4 \end{split} \]

Substitute:

\[ 9[(x-3)^2 - 9] + 4[(y+2)^2 - 4] = -61 \]

Expand:

\[ 9(x-3)^2 - 81 + 4(y+2)^2 - 16 = -61 \]

Combine constants:

\[ 9(x-3)^2 + 4(y+2)^2 = 36 \]

Divide by 36:

\[ \frac{(x-3)^2}{4} + \frac{(y+2)^2}{9} = 1 \]

This ellipse has center \((3,-2)\) and vertical major axis.

WarningCommon Mistakes
  • Mixing up whether \(a^2\) or \(b^2\) is larger
  • Forgetting the relationship \(c^2 = a^2 - b^2\)
  • Putting \(a^2\) under the wrong variable when re-writing
  • Forgetting to divide all terms when converting to standard form
  • Treating an ellipse like a circle when denominators look “almost” equal

Practice Problems

  1. Find the foci of
    \[ \frac{(x+1)^2}{36} + \frac{(y-4)^2}{4} = 1 \]

  2. Write the equation of an ellipse with center \((5,-3)\), horizontal major axis, \(a=6\), \(b=2\).

  3. Convert the ellipse
    \[ 4x^2 + 9y^2 - 16x + 36y + 64 = 0 \]
    to standard form.

  4. Identify whether the ellipse has a horizontal or vertical major axis:
    \[ \frac{(x-2)^2}{9} + \frac{(y+3)^2}{49} = 1 \]

  5. A vertical ellipse has center \((0,0)\) and foci at \((0,\pm 5)\). If \(b=12\), find the equation.

TipStep-by-Step Solutions

1. \(a^2=36\), \(b^2=4\)
\(c^2 = 36 - 4 = 32\)\(c = 4\sqrt{2}\)
Foci: \((-1 \pm 4\sqrt{2},\; 4)\)

2. Horizontal major axis → \(a^2=36\) under \((x-5)^2\)
\(b^2=4\) under \((y+3)^2\)
Equation:
\[ \frac{(x-5)^2}{36} + \frac{(y+3)^2}{4} = 1 \]

3. Standard form:
\[ \frac{(x-2)^2}{9} + \frac{(y+2)^2}{4} = 1 \]
Center \((2,-2)\)

4. Larger denominator is 49 → vertical major axis.

5. Foci at \((0,\pm 5)\)\(c=5\)
Use \(c^2 = a^2 - b^2\): \[ a^2 = c^2 + b^2 = 25 + 144 = 169 \]
Vertical:
\[ \frac{x^2}{144} + \frac{y^2}{169} = 1 \]

Summary

  • Ellipses stretch circles into two unequal axes.
  • Larger denominator → major axis direction.
  • Use \(c^2 = a^2 - b^2\) to find foci distance.
  • ACT problems frequently involve identifying orientation, converting forms, and writing equations from geometric facts.
TipQuick Tips
  • Always identify \(a\), \(b\), and orientation before doing anything else.
  • Center shifts are easy to overlook—double-check signs.
  • If you’re converting from general form: group, complete the square, divide.