Section 5.2 Ellipse

An ellipse is the set of points in a plane the sum of whose distances from two fixed points \(F_1\) and \(F_2\) (the foci) is constant. For \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) with \(a \gt b\text{,}\) the segment joining \((a,0)\) and \((-a,0)\) is the semi-major axis, the segment joining \((0,b)\) and \((0,-b)\) is the semi-minor axis, and the foci satisfy \(c^2 = a^2 - b^2 \gt 0\text{.}\)

Diagram Exploration Keyboard Controls

Key	Action
Enter, A	Activate keyboard driven exploration
B	Activate menu driven exploration
Escape	Leave exploration mode
Cursor down	Explore next lower level
Cursor up	Explore next upper level
Cursor right	Explore next element on level
Cursor left	Explore previous element on level
X	Toggle expert mode
W	Extra details if available
Space	Repeat speech
M	Activate step magnification
Comma	Activate direct magnification
N	Deactivate magnification
Z	Toggle subtitles
C	Cycle contrast settings
T	Monochrome colours
L	Toggle language (if available)
K	Kill current sound
Y	Stop sound output
O	Start and stop sonification
P	Repeat sonification output

Figure 5.7. An ellipse with foci \(F_1(c,0)\) and \(F_2(-c,0)\text{.}\) Each slanted segment from \((0,b)\) to a focus has length \(a\text{.}\)

Figure 5.8. Animation: as \(P\) travels around the ellipse, the sum \(r_1 + r_2\) of its distances to the two foci stays fixed at \(2a = 6\text{.}\)

Where do the formulas for \(a\) come from? Two facts do all the work. First, evaluating the definition at the vertex \((a,0)\) shows that the two focal radii there are \(a-c\) and \(a+c\text{,}\) so their sum is

\begin{equation*} r_1 + r_2 = (a-c) + (a+c) = 2a, \end{equation*}

which identifies the constant in the definition as \(2a\text{;}\) equivalently \(a = \frac{1}{2}(r_1 + r_2)\text{.}\) Second, evaluating at the co-vertex \((0,b)\) forces \(r_1 = r_2 = a\) by symmetry, and that radius is the hypotenuse of a right triangle with legs \(b\) and \(c\text{.}\) Pythagoras then gives

\begin{equation*} a^2 = b^2 + c^2 \qquad\Longleftrightarrow\qquad c^2 = a^2 - b^2. \end{equation*}

So for an ellipse \(a\) is the hypotenuse of that triangle, which is why \(a\) is the largest of the three lengths and why \(c^2 = a^2 - b^2\) carries a minus sign.

Figure 5.9. Animation: deriving the formula for \(a\text{.}\) Sliding \(P\) to the vertex shows the constant equals \(2a\text{;}\) sliding it to the co-vertex builds the right triangle giving \(a^2 = b^2 + c^2\text{.}\)

Figure 5.10. Animation: in practice, put the equation in standard form and read \(a^2\) off the larger denominator; then \(a\) is the semi-major axis, measured from the center to a vertex.

Example 5.11. Horizontal ellipse.

Sketch the graph of \(4x^2 + 9y^2 = 36\) and locate the foci.

Solution.

Dividing through by \(36\) puts the equation in standard form:

\begin{equation*} \frac{x^2}{9} + \frac{y^2}{4} = 1. \end{equation*}

The larger denominator sits under \(x^2\text{,}\) so the major axis is horizontal with \(a = 3\) and \(b = 2\text{.}\) Then

\begin{equation*} c^2 = a^2 - b^2 = 9 - 4 = 5 \;\Longrightarrow\; c = \sqrt{5}, \end{equation*}

so the foci are \(F_1(\sqrt{5},0)\) and \(F_2(-\sqrt{5},0)\text{.}\)

Diagram Exploration Keyboard Controls

Key	Action
Enter, A	Activate keyboard driven exploration
B	Activate menu driven exploration
Escape	Leave exploration mode
Cursor down	Explore next lower level
Cursor up	Explore next upper level
Cursor right	Explore next element on level
Cursor left	Explore previous element on level
X	Toggle expert mode
W	Extra details if available
Space	Repeat speech
M	Activate step magnification
Comma	Activate direct magnification
N	Deactivate magnification
Z	Toggle subtitles
C	Cycle contrast settings
T	Monochrome colours
L	Toggle language (if available)
K	Kill current sound
Y	Stop sound output
O	Start and stop sonification
P	Repeat sonification output

Figure 5.12. The ellipse \(\frac{x^2}{9}+\frac{y^2}{4}=1\) with foci at \(\left(\pm\sqrt5,\,0\right)\text{.}\)

Example 5.13. A vertical ellipse (foci on the \(y\)-axis).

Sketch the graph of \(25x^2 + 16y^2 = 400\) and locate the foci.

Solution.

Dividing by \(400\text{,}\)

\begin{equation*} \frac{x^2}{16} + \frac{y^2}{25} = 1. \end{equation*}

This time the larger denominator, \(25\text{,}\) sits under \(y^2\text{,}\) so the major axis is vertical. The semi-major axis is \(a = \sqrt{25} = 5\) (along the \(y\)-axis) and the semi-minor axis is \(b = \sqrt{16} = 4\) (along the \(x\)-axis). Hence

\begin{equation*} c^2 = a^2 - b^2 = 25 - 16 = 9 \;\Longrightarrow\; c = 3. \end{equation*}

Because the major axis is vertical, the foci lie on the \(y\)-axis: \(F_1(0,3)\) and \(F_2(0,-3)\text{.}\) The lesson of the twist: read off the larger denominator first, since it decides which axis carries the foci.

Diagram Exploration Keyboard Controls

Key	Action
Enter, A	Activate keyboard driven exploration
B	Activate menu driven exploration
Escape	Leave exploration mode
Cursor down	Explore next lower level
Cursor up	Explore next upper level
Cursor right	Explore next element on level
Cursor left	Explore previous element on level
X	Toggle expert mode
W	Extra details if available
Space	Repeat speech
M	Activate step magnification
Comma	Activate direct magnification
N	Deactivate magnification
Z	Toggle subtitles
C	Cycle contrast settings
T	Monochrome colours
L	Toggle language (if available)
K	Kill current sound
Y	Stop sound output
O	Start and stop sonification
P	Repeat sonification output

Figure 5.14. The vertical ellipse \(\frac{x^2}{16}+\frac{y^2}{25}=1\text{;}\) the foci \((0,\pm 3)\) sit on the major axis, and each segment from \((4,0)\) to a focus has length \(a = 5\text{.}\)

You have attempted of activities on this page.

Prev Top Next