7.1 Ellipses

If you were to ask a random person “What is a circle?” a typical response would be to kick the can down the road: “Something that’s round.” There is a simple definition:

Definition 7.1

A circle is the set of all points in a plane that are a fixed distance from a fixed point in that plane. The fixed point is the center of the circle, while the fixed distance from the center is the circle’s radius.

Similarly, the question “What is an ellipse?” would likely be answered with “an oval,” “something egg-shaped,” or “a squished circle.” A precise definition would be:

Definition 7.2

An ellipse is the set of all points in a plane such that the sum of the distances from each of those points to two fixed points in the plane is the same constant. The two fixed points are the foci—plural of focus—of the ellipse.

Figure 7.1.1 illustrates the above definitions, with a point $P$ moving along each curve.

tikzpicture[every node/.style=font=] [dashed] (0,0) -- (70:1.2) node[pos=0.4,above left] r; [linecolor,line width=1.5pt] (0,0) circle (1.2); [above] at (70:1.2) P; [below] at (0,-0.1) C; (70:1.2) circle (2.5pt); (0,0) circle (2.5pt); (-3.1,0) -- (3.1,0); tikzpicture — Figure 7.1.1 The circle and its “squished” sibling the ellipse

tikzpicture[every node/.style=font=] [line width=1.5pt,linecolor] (0,0) ellipse (2.2 and 1.2); (75:2.2 and 1.2) circle (2.5pt); [above right] at (75:2.2 and 1.2) P; [dashed] (-1.1,0) -- (75:2.2 and 1.2) node[pos=0.35,above left] d_1; [dashed] (1.1,0) -- (75:2.2 and 1.2) node[pos=0.25,above right] d_2; (-1.1,0) circle (2.5pt); [below] at (-1.1,-0.1) F_1; (1.1,0) circle (2.5pt); [below] at (1.1,-0.1) F_2; (-3.6,0) -- (3.6,0); tikzpicture — Figure 7.1.1 The circle and its “squished” sibling the ellipse

Along the ellipse the sum $d_1+d_2$ of the distances from $P$ to the foci remains constant. The circle’s definition makes it easy to imagine its shape, especially for anyone who has drawn a circle with a compass. The definition of the ellipse, on the other hand, might not immediately suggest an “oval” shape. Its shape becomes apparent when physically constructing an ellipse by hand, using only the definition. Stick two pins in a board and tie the ends of a piece of string to the pins, with the string long enough so that there is some slack (see Figure 7.1.2(a)). The pins will be the foci of the ellipse.

tikzpicture[>=latex,every node/.style=font=] [line width=2pt] (-3,-1.8) rectangle (3,1.8); [very thick,double] (-1.1,0) .. controls (-0.5,-1.7) and (0.1,1.95) .. (1.1,0); [shading=ball] (-1.1,0) circle (4pt); [shading=ball] (1.1,0) circle (4pt); tikzpicture — Figure 7.1.2 Construction of an ellipse

tikzpicture[>=latex,every node/.style=font=] [line width=2pt] (-3,-1.8) rectangle (3,1.8); [dashed] (0,0) ellipse (2.2 and 1.13); [rounded corners,very thick,double] (-1.1,0) -- (75:2.2 and 1.2) -- (1.1,0); [rotate=-50] at (78:2.2 and 1.32) ; [shading=ball] (-1.1,0) circle (4pt); [shading=ball] (1.1,0) circle (4pt); tikzpicture — Figure 7.1.2 Construction of an ellipse

Pull the string taut with a pencil whose point is touching the board, then move the pencil around as far as possible on all sides of the pins. The drawn figure will be an ellipse, as in Figure 7.1.2(b). The length of the string is the constant sum $d_1+d_2$ of distances from points on the ellipse to the foci. The symmetry of the ellipse is obvious.

There is some terminology connected with ellipses. The principal axis is the line containing the foci, and the center is halfway between the foci, as in Figure 7.1.3:

tikzpicture[>=latex,every node/.style=font=] [green!60!blue!70,line width=1.5pt] (0,1.6) -- (0,-1.6); [linecolor,line width=1.5pt] (0,0) ellipse (4 and 1.6); [red,dashed,line width=1pt] (-6.7,0) -- (6.7,0) node[pos=0.93,above,align=center,black] principal\ ; [red!60,line width=1.5pt] (-4,0) -- (4,0); [red!60,decorate, decoration=brace,raise=5pt,amplitude=10pt,aspect=0.7] (-3.9,0) -- (3.9,0) node[black,pos=0.7,above,outer ysep=14pt] major axis; [green!60!blue!70,decorate, decoration=brace,raise=7pt,amplitude=10pt,aspect=0.7,mirror] (0,1.5) -- (0,-1.5) node[black,pos=0.7,left,outer xsep=16pt,align=right] minor\ ; (-2.6,0) circle (2.5pt); (2.6,0) circle (2.5pt); (0,0) circle (2.5pt); (-4,0) circle (2.5pt); (4,0) circle (2.5pt); [below,align=center] at (-2.6,-0.1) F_1\\(focus); [below,align=center] at (2.6,-0.1) F_2\\(focus); [below right] at (0,-0.1) C (center); [below left,align=right] at (-4,-0.1) V_1\\(vertex); [below right,align=left] at (4,-0.1) V_2\\(vertex); tikzpicture — Figure 7.1.3 The parts of an ellipse

The vertexes are the points where the ellipse intersects the principal axis. The major axis is the chord joining the vertexes, and the minor axis is the chord through the center that is perpendicular to the major axis. The two semi-major axes are the halves of the major axis joining the center to the vertexes ($\overline{CV_1}$ and $\overline{CV_2}$ in Figure 7.1.3). Likewise the semi-minor axes are the two halves of the minor axis. A chord through the center is a diameter. Notice that a circle is the special case of an ellipse with identical foci (i.e. the foci and center are all the same point). Ellipses appear in nature (e.g. the orbits of planets around the Sun) and in many applications. The ancient Greeks were able to derive many properties of the ellipse from its purely geometric definition.^[1] Nowadays those properties are typically derived using methods from analytic geometry—the study of geometric objects in the context of coordinate systems.^[2] You have already seen the equation of an ellipse in the $xy$-plane centered at the origin: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a>b>0$, with the $x$-axis as the principal axis. The equation is straightforward to derive from the definition of the ellipse.

[r] In the $xy$-plane, let the foci of an ellipse be the points $(\pm c,0)$ for some $c>0$, so that the center is the origin $(0,0)$ and the $x$-axis is the principal axis, as in the figure on the right. Denote by $2a$ the constant sum $d_1+d_2$ of the distances from points $(x,y)$ on the ellipse to the foci, with $a > 0$. Notice that $a>c$, since the distance $2c$ between the foci must be less than $d_1+d_2 = 2a$. Then by the distance formula,

\begin{align*} d_1 ~+~ d_2 ~&=~ 2a\\ \sqrt{(x+c)^2 + y^2} ~+~ \sqrt{(x-c)^2 + y^2} ~&=~ 2a\\ \left(\sqrt{(x-c)^2 + y^2}\right)^2 ~&=~ \left(2a ~-~ \sqrt{(x+c)^2 + y^2}\right)^2\\ (x-c)^2 ~+~ \cancel{y^2} ~&=~ 4a^2 ~-~ 4a\,\sqrt{(x+c)^2 + y^2} ~+~ (x+c)^2 ~+~ \cancel{y^2}\\ 4a\,\sqrt{(x+c)^2 + y^2} ~&=~ 4a^2 ~+~ (x+c)^2 ~-~ (x-c)^2\\ \cancel{4}a\,\sqrt{(x+c)^2 + y^2} ~&=~ \cancel{4}a^2 ~+~ \cancel{4}xc\\ \sqrt{(x+c)^2 + y^2} ~&=~ a ~+~ \tfrac{c}{a}x\\ x^2 ~+~ \cancel{2cx} ~+~ c^2 ~+~ y^2 ~&=~ a^2 ~+~ \cancel{2cx} ~+~ \tfrac{c^2}{a^2}x^2\\ \left(1 - \tfrac{c^2}{a^2}\right)\,x^2 ~+~ y^2 ~&=~ a^2 ~-~ c^2\\[4pt] \frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} ~&=~ 1\\[4pt] \frac{x^2}{a^2} + \frac{y^2}{b^2} ~&=~ 1 \quad\text{where $b^2 = a^2 - c^2$ (and so $a > b > 0$)}\quad\checkmark \end{align*}

tikzpicture[>=latex,every node/.style=font=] [->,black!60,line width=1pt] (-2.8,0) -- (3,0) node[right] x; [->,black!60,line width=1pt] (0,-1.8) -- (0,2) node[above] y; [line width=1.5pt,linecolor] (0,0) ellipse (2.2 and 1.2); (-1.1,0) circle (2.5pt); [below] at (-1.1,-0.1) -c; (1.1,0) circle (2.5pt); [below] at (1.1,-0.1) c; (-2.2,0) circle (2.5pt); [below left] at (-2.2,-0.1) -a; (2.2,0) circle (2.5pt); [below right] at (2.2,-0.1) a; [below right] at (0,-0.1) 0; (0,1.2) circle (2.5pt); [above left] at (0,1.2) b; (0,-1.2) circle (2.5pt); [below left] at (0,-1.2) -b; [above] at (2,1) x^2a^2 + y^2b^2 = 1; tikzpicture — Figure 7.1.4

The graph of the resulting ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a>b>0$ and foci at $(\pm c,0)$ is shown in Figure 7.1.4. Since the the $x$-axis is principal axis then the vertexes are found by setting $y=0$: $x=\pm a$. The vertexes are thus $(\pm a,0)$, so the major axis goes from $(-a,0)$ to $(a,0)$ and has length $2a$ (i.e. the semi-major axis has length $a$). Similarly, setting $x=0$ shows the minor axis goes from $(0,-b)$ to $(0,b)$, (i.e. the semi-minor axis has length $b$). Since $a>b>0$ and $b^2 = a^2 - c^2$,
then $c = \sqrt{a^2 - b^2}$. Thus, for any ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a>b>0$, the foci will be at $(\pm c,0)$, where $c = \sqrt{a^2 - b^2}$. The foci can be used to define an important geometric property of the ellipse:

Definition 7.3

The eccentricity of an ellipse is the ratio of the distance between the foci to the length of the major axis. In the case of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a>b>0$, the eccentricity $e$ is given by

\[ e ~=~ \frac{c}{a} ~=~ \frac{\sqrt{a^2 - b^2}}{a} ~. \]

The eccentricity $e$ is a measure of how “oval” an ellipse is, with $0<e<1$. The boundary case $e=0$ is a circle, while $e=1$ is a line segment; an ellipse is somewhere in between—the closer $e$ gets to 1, the more “squished” the ellipse. See Figure 7.1.5.

tikzpicture[every node/.style=font=] [linecolor,line width=1.5pt] (0,0) circle (1); (0,0) circle (2.5pt); [below] at (0,-0.1) C; (-1.5,0) -- (1.5,0); tikzpicture — Figure 7.1.5 Eccentricity $e$

tikzpicture[every node/.style=font=] [line width=1.5pt,linecolor] (0,0) ellipse (1.15 and 1); (-0.58,0) circle (2.5pt); [below] at (-0.58,-0.1) F_1; (0.58,0) circle (2.5pt); [below] at (0.58,-0.1) F_2; (-1.5,0) -- (1.5,0); tikzpicture — Figure 7.1.5 Eccentricity $e$

Earth’s elliptical orbit around the Sun is almost circular: the eccentricity is 0.017. Only the orbits of Venus and Neptune (both at 0.007) have a lower eccentricity among the nine planets in the solar system, while Pluto’s (0.252) has the highest.

It is left as an exercise to show that the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a>b>0$ can be written in terms of the eccentricity $e$:

\begin{equation} y^2 ~=~ (1 - e^2)\,(a^2 - x^2) \tag{7.1}\end{equation}

Example 7.1

Find the area inside the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Solution: By symmetry the area will be four times the area in the first quadrant. Solving for $y$ in the equation of the ellipse gives

\[ y^2 ~=~ b^2 ~-~ \frac{b^2 x^2}{a^2} \quad\Rightarrow\quad y ~=~ b\,\sqrt{1 - \frac{x^2}{a^2}} ~=~ \frac{b}{a}\,\sqrt{a^2 - x^2} \]

for the upper hemisphere. Thus,

\begin{align*} \text{Area} ~&=~ 4\,\int_0^a y~\dx ~=~ \frac{4b}{a}\,\int_0^a \sqrt{a^2 - x^2}~\dx\\[6pt] &=~ \frac{4b}{a}\,\left(\frac{a^2}{2}\,\sin^{-1} \left(\frac{x}{a}\right) ~+~ \frac{1}{2}\,x\,\sqrt{a^2 - x^2} ~\Biggr|_0^a\right) \quad\text{(by formula (6.7) in Section 6.3)}\\[6pt] &=~\frac{4b}{a}\,\left(\frac{a^2}{2}\,\frac{\pi}{2} ~+~ 0 ~-~ (0 ~+~ 0)\right)\\[6pt] &=~ \pi\,ab \end{align*}

tikzpicture[>=latex, every node/.style=font=, decoration=markings,mark=at position 0.55 with triangle 60] [line width=1.5pt,linecolor] (0,0) ellipse (2.2 and 1.2); (75:2.2 and 1.2) circle (2.5pt); [above right] at (75:2.2 and 1.2) P; [postaction=decorate,dashed] (-1.1,0) -- (75:2.2 and 1.2); [postaction=decorate,dashed] (75:2.2 and 1.2) -- (1.1,0); (-1.1,0) circle (2.5pt); [below] at (-1.1,-0.1) F_1; (1.1,0) circle (2.5pt); [below] at (1.1,-0.1) F_2; tikzpicture — Figure 7.1.6

A remarkable feature of the ellipse is the reflection property: light shone from one focus to any point on the ellipse will reflect to the other focus. Figure 7.1.6 shows light emanating from the focus $F_1$ and reflecting off a point $P$ on the ellipse to the other focus $F_2$. Fermat’s Principle from Example 4.4 in Section 4.1 showed that the incoming angle $\theta_1$ (angle of incidence) of light from point A will equal the outgoing angle $\theta_2$ (angle of reflection) to point B for light reflecting off a flat reflective surface at point $P$, as in Figure 7.1.7(a). Fermat’s Principle also applies to curved surfaces—e.g. an ellipse—with the angles measured relative to the tangent line to the curve at the point of reflection, as in Figure 7.1.7(b).

tikzpicture[every node/.style=font=] (180:0.9) arc (180:135:0.9); (0:0.9) arc (0:45:0.9); [linecolor,line width=1.5pt] (-2,2) -- (0,0) -- (1.5,1.5); [linecolor,line width=1pt,-triangle 60] (-2,2) -- (-1,1); [linecolor,line width=1pt,-triangle 60] (0,0) -- (1,1); [line width=2pt] (-2.5,0) -- (2.5,0); [above] at (-2,2.1) A; [above] at (1.5,1.6) B; [below] at (0,-0.1) P; at (160:0.6) _1; at (20:0.6) _2; (-2,2) circle (2.5pt); (1.5,1.5) circle (2.5pt); tikzpicture — Figure 7.1.7 Fermat’s Principle for reflection: $\theta_1 = \theta_2$

tikzpicture[every node/.style=font=] (180:0.9) arc (180:135:0.9); (0:0.9) arc (0:45:0.9); [linecolor,line width=1.5pt] (-1.5,1.5) -- (0,0) -- (1.2,1.2); [linecolor,line width=1pt,-triangle 60] (-1.5,1.5) -- (-1,1); [linecolor,line width=1pt,-triangle 60] (0,0) -- (0.95,0.95); [dashed] (-2.5,0) -- (2.5,0) node[above,align=center] tangent\ ; [line width=2pt] (-2,-1) parabola[bend pos=0.5] bend +(0,1) (2,-1); [above] at (-1.5,1.6) A; [above] at (1.2,1.3) B; [below] at (0,-0.1) P; at (160:0.6) _1; at (20:0.6) _2; (-1.5,1.5) circle (2.5pt); (1.2,1.2) circle (2.5pt); tikzpicture — Figure 7.1.7 Fermat’s Principle for reflection: $\theta_1 = \theta_2$

Notice that Fermat’s Principle is equivalent to saying that the angles $\alpha_1$ and $\alpha_2$ that the light’s path makes with the normal line through the point of reflection are equal, since each angle would equal $90\Degrees - \theta$, as in Figure 7.1.8(a):

tikzpicture[every node/.style=font=] (0:0.9) arc (0:180:0.9); [linecolor,line width=1.5pt] (-1.2,1) -- (0,0) -- (1.2,1); [linecolor,line width=1pt,-triangle 60] (-1.2,1) -- (-0.72,0.6); [linecolor,line width=1pt,-triangle 60] (0,0) -- (0.96,0.8); [dashed] (-2.2,0) -- (2.2,0) node[above,align=center] tangent\ ; [dashed] (0,0) -- (0,1.5) node[above] normal line; [line width=2pt] (-2,-0.7) parabola[bend pos=0.5] bend +(0,0.7) (2,-0.7); [above] at (-1.2,1.1) A; [above] at (1.2,1.1) B; [below] at (0,-0.1) P; at (160:0.6) ; at (20:0.6) ; at (115:0.6) _1; at (65:0.6) _2; (-1.2,1) circle (2.5pt); (1.2,1) circle (2.5pt); tikzpicture — Figure 7.1.8 Fermat’s Principle with normal line

tikzpicture[>=latex,every node/.style=font=] [linecolor,line width=1.5pt] (3.5,0) arc[start angle=0,end angle=180, x radius=3.5cm,y radius=1.6cm]; [->,black!60,line width=1pt] (-3.8,0) -- (4,0) node[right] x; [->,black!60,line width=1pt] (0,-0.4) -- (0,2) node[above] y; (-2.1,0) -- (70:3.5 and 1.6) -- (2.1,0); (-2.1,0) circle (2.5pt); (70:3.5 and 1.6) circle (2.5pt); (2.1,0) circle (2.5pt); (0.947,0) circle (2.5pt); (-3.5,0) circle (2.5pt); (3.5,0) circle (2.5pt); [above left] at (-2.1,0.1) F_1; [below] at (-2.1,-0.1) -c; [above right] at (2.1,0.1) F_2; [below] at (2.1,-0.1) c; [below left] at (0,-0.1) 0; [below] at (-3.5,-0.1) -a; [below] at (3.5,-0.1) a; [below left] at (0,1.6) b; [below] at (0.947,-0.1) N; [above right] at (70:3.5 and 1.6) P(x_0,y_0); (0.947,0) -- ++(80.55:2.1) node[above right] n = normal line; (70:3.5 and 1.6) -- ++(170.55:2) node[above left] tangent line; (70:3.5 and 1.6) -- ++(170.55:-2); (70:3.5 and 1.6) -- +(260.55:1) arc(260.55:301:1); (70:3.5 and 1.6) -- +(260.55:1) arc(260.55:204.5:1); at (0.8,0.93) _1; at (1.35,0.75) _2; tikzpicture — Figure 7.1.8 Fermat’s Principle with normal line

Thus, to prove the reflection property, it suffices to prove that the normal line $n$ to the ellipse at $P$ bisects the angle $\angle F_1PF_2$ in Figure 7.1.8(b)—this would make $\alpha_1=\alpha_2$, so that the indicated path from $F_1$ to $P$ to $F_2$ satisfies Fermat’s Principle. First, let $P=(x_0,y_0)$ be a point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, with $a>b>0$. Assume that $P$ is not a vertex (i.e. $y_0 \ne 0$), otherwise the reflection property holds trivially. By Exercise 15 in Section 3.4, the equation of the tangent line to the ellipse at $P=(x_0,y_0)$ is

\begin{equation} \frac{x x_0}{a^2} ~+~ \frac{y y_0}{b^2} ~=~ 1 ~, \tag{7.2}\end{equation}

so that its slope is $-\frac{b^2 x_0}{a^2 y_0}$. Hence, the negative reciprocal $\frac{a^2 y_0}{b^2 x_0}$ is the slope of the normal line $n$, whose equation—valid even when $x_0=0$ (i.e. when $y_0=\pm b$)— is then

\begin{equation} b^2 x_0\,(y - y_0) ~=~ a^2 y_0\,(x - x_0) ~. \tag{7.3}\end{equation}

Setting $y=0$ and solving for $x$ shows the $x$-intercept of $n$ is at

\[ x ~=~ \frac{(a^2 - b^2)\,x_0 y_0}{a^2 y_0} ~=~ \frac{c^2}{a^2}\,x_0 ~=~ e^2\,x_0 \]

Let $N = (e^2 x_0,0)$ be that $x$-intercept, as in Figure 7.1.8(b). The distance $F_1N$ from the focus $F_1 =(-c,0)=(-ea,0)$ to $N$ is then

\[ F_1N ~=~ e^2\,x_0 ~-~ (-ea) ~=~ e\,(a + ex_0) \]

while the distance $F_2N$ from the focus $F_2 =(c,0)=(ea,0)$ to $N$ is

\[ F_2N ~=~ ea ~-~ e^2\,x_0 ~=~ e\,(a - ex_0) ~. \]

Therefore,

\[ \frac{F_1N}{F_2N} ~=~ \frac{e\,(a + ex_0)}{e\,(a - ex_0)} ~=~ \frac{a + ex_0}{a - ex_0} ~. \]

By the distance formula, the distance $F_1P$ from $F_1=(-ea,0)$ to $P=(x_0,y_0)$ is given by

\begin{align*} (F_1P)^2 ~&=~ (x_0 + ea)^2 ~+~ y_0^2\\ &=~ x_0^2 ~+~ 2eax_0 ~+~ e^2a^2 ~+~ (1-e^2)\,(a^2 - x_0^2) \quad\text{(by formula (7.1))}\\ (F_1P)^2 ~&=~ a^2 ~+~ 2eax_0 ~+~ e^2x_0^2 ~=~ (a + ex_0)^2\\ F_1P ~&=~a + ex_0 ~. \end{align*}

Similarly, the distance $F_2P$ from $F_2=(ea,0)$ to $P=(x_0,y_0)$ is given by

\begin{align*} (F_2P)^2 ~&=~ (x_0 - ea)^2 ~+~ y_0^2 ~=~ x_0^2 ~-~ 2eax_0 ~+~ e^2a^2 ~+~ (1-e^2)\,(a^2 - x_0^2)\\ (F_2P)^2 ~&=~ a^2 ~-~ 2eax_0 ~+~ e^2x_0^2 ~=~ (a - ex_0)^2\\ F_2P ~&=~a - ex_0 ~. \end{align*}

[r] Thus,

\[ \frac{F_1P}{F_2P} ~=~ \frac{a + ex_0}{a - ex_0} ~=~ \frac{F_1N}{F_2N} ~, \]

which means that $\alpha_1 = \alpha_2$:^[3] by the Law of Sines, and with $\theta =\angle F_2NP$ as in the figure on the right,

\[ \frac{\sin\,\alpha_2}{F_2N} ~=~ \frac{\sin\,\theta}{F_2P} ~=~ \frac{\sin\,(180\Degrees - \theta)}{F_2P} ~=~ \frac{\sin\,(180\Degrees - \theta)}{F_1P}\,\cdot\,\frac{F_1P}{F_2P} ~=~ \frac{\sin\,\alpha_1}{F_1N}\,\cdot\,\frac{F_1N}{F_2N} ~=~ \frac{\sin\,\alpha_1}{F_2N} \]

and thus $\sin\,\alpha_2 = \sin\,\alpha_1$, so that $\alpha_2 = \alpha_1$ (since $0\Degrees < \alpha_1,\,\alpha_2 < 90\Degrees$). $\quad\checkmark$

tikzpicture[>=latex,every node/.style=font=] [->,black!60,line width=1pt] (-1.8,0) -- (2,0) node[right] x; [->,black!60,line width=1pt] (0,-2.6) -- (0,3) node[right] y; [line width=1.5pt,linecolor] (0,0) ellipse (1.2 and 2.2); (0,-1.1) circle (2.5pt); [left] at (-0.1,-1.1) -c; (0,1.1) circle (2.5pt); [left] at (-0.1,1.1) c; (-1.2,0) circle (2.5pt); [below left] at (-1.2,-0.1) -b; (1.2,0) circle (2.5pt); [below right] at (1.2,-0.1) b; [below right] at (0,-0.1) 0; (0,2.2) circle (2.5pt); [above left] at (0,2.2) a; (0,-2.2) circle (2.5pt); [below left] at (0,-2.2) -a; [above] at (1.3,2) x^2b^2 + y^2a^2 = 1; tikzpicture — Figure 7.1.9

An ellipse of the form

\[ \frac{x^2}{b^2} ~+~ \frac{y^2}{a^2} ~=~ 1 \]

with $a>b>0$ simply switches the roles of $x$ and $y$ in the previous examples: the principal axis is now the $y$-axis, the foci are at $(0,\pm c)$, where $c = \sqrt{a^2 - b^2}$, and the vertexes are at $(0,\pm a)$, as in Figure 7.1.9.

Thus, the largest denominator on the left side of an equation of the form $\frac{x^2}{\square^2} + \frac{y^2}{\square^2} = 1$ tells you which axis is the principal axis. For example, the principal axis of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is the $x$-axis (since $25>16$), while the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ has the $y$-axis as its principal axis (since $9>4$).

Exercises

Construct an ellipse using the procedure shown in Figure 7.1.2. Place the two pins 7in apart and use a 10in piece of string.

For Exercises 2-6, sketch the graph of the given ellipse, indicate the major and minor axes and exact locations of the foci and vertexes, and find the eccentricity $e$.

$\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$
$\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$
$\dfrac{4x^2}{25} + \dfrac{y^2}{4} = 1$
$x^2 + 4y^2 = 1\vphantom{\dfrac{x^2}{15}}$
$25x^2 + 9y^2 = 225\vphantom{\dfrac{x^2}{15}}$
Show that for $a>b>0$ the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$ with eccentricity $e$ can be written as $y^2 = (1-e^2)\,(a^2 - x^2)$.
Use Example 7.1 to show the area inside the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentricity $e$ is $\pi a^2\,\sqrt{1-e^2}$.
For all $a>b>0$, find the points of intersection of the ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.
Show that the vertexes are the closest and farthest points on an ellipse to either focus.

Show that any line of slope $m$ that is tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ must be of the form

\[ y ~=~ mx ~\pm~ \sqrt{a^2 m^2 \;+\; b^2} ~. \]
A 10ft ladder with a mark 3ft from the top rests against a wall. If the top of the ladder slides down the wall, with the foot of the ladder sliding away from the wall on the ground, as in Figure 7.1.10, then show the mark moves along part of an ellipse.

Figure 7.1.10 Exercise 12

Figure 7.1.11 Exercise 13

Figure 7.1.12 Exercise 14
Another definition of an ellipse is the set of points $P$ for which the ratio of the distance from $P$ to a fixed point $F$ (a focus) to the distance from $P$ to a fixed line $D$ (the directrix) is a constant $e<1$ (the eccentricity): $\frac{PF}{PG}=e$, as in Figure 7.1.11. Use this definition to show that the equation of an ellipse with focus $(c,0)$ can be written as $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ for some $a>b>0$. Find the equation of the directrix.
Show that the set of intersection points of all perpendicular tangent lines to an ellipse form a circle, as in Figure 7.1.12 (showing two such tangent lines $T_1 \perp T_2$).
A chord of an ellipse that passes through a focus and is perpendicular to the major axis is a latus rectum. Show that for the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$ with $a>b>0$ the length of each latus rectum is $\frac{2b^2}{a}$.
Suppose that the normal line at one end of a latus rectum of an ellipse passes through an end of the minor axis. Show that the eccentricity $e$ is a root of the equation $e^4 + e^2 - 1 = 0$, then find $e$.
Show that the set of all midpoints of a family of parallel chords in an ellipse lie on a diameter. (Hint: Use symmetry with chords of slope $m \ne 0$.)

The word ellipse is in fact due to the Greek astronomer and geometer Apollonius of Perga (ca. 262-190 B.C.), which seems an improvement over the name “thyreos” that Euclid (ca. 360-300 B.C.) had given the shape. ↩
Pioneered by the French mathematician and philosopher René Descartes (1596-1650), for whom the Cartesian coordinate system is named. The proposition “I think, therefore I am” (Cogito, ergo sum) is due to Descartes. ↩
This follows directly from Proposition 3 in Book VI of Euclid’s Elements. See the purely geometric proof on pp.125-126 in Euclid, Elements, (Thomas L. Heath translation), Santa Fe, NM: Green Lion Press, 2002. ↩