7.3: Hyperbolas

In the previous two sections you have seen curves with eccentricity $e=0$ (circles), $0<e<1$ (ellipses) and $e=1$ (parabolas). The remaining case is $e>1$: the hyperbola, whose definition is similar to the second definition of the ellipse.

It will be shown in Section 7.4 that the curve $y=\frac{1}{x}$ is a hyperbola, which has two branches (see Figure [fig:hyper1x]). In general a hyperbola resembles a “wider” or less “cupped” parabola, and it has two symmetric branches (and hence two foci and two directrices) as well as two asymptotes.

The ratio of distances referred to in the definition of the hyperbola also appears in the second definition of the ellipse (where the ratio is smaller than 1) and in the definition of the parabola (where the ratio equals 1). In all three cases that ratio is the eccentricity. See Figure [fig:ecccompare](a) for the comparisons.

There is an analogue for Figure [fig:ecccompare](a) in terms of orbits. For example, an object approaching the Sun must meet or exceed the escape velocity to overcome the Sun’s gravitational pull and avoid returning to orbit. Figure [fig:ecccompare](b) shows the three possible trajectories—hyperbola, parabola and ellipse—in terms of the object’s velocity $v$ and the escape velocity $v_E$. Note the apparent correlation between the eccentricity of the object’s path and its speed as a fraction of the escape velocity (i.e. $\frac{v}{v_E}$). Figure [fig:hyperbolavert] illustrates the definition of a hyperbola, consisting of points $P$ whose distance $PF$ from the focus $F$ exceeds the distance $PG$ to the directrix $D$ in a way so that the ratio $\frac{PF}{PG}$ is always the same constant $e>1$ (the eccentricity).

To derive the equation of a hyperbola with eccentricity $e>1$, assume the focus is on the $x$-axis at $(ea,0)$, with $a>0$, and the line $x=\frac{a}{e}$ is the directrix, as in Figure [fig:hypereqn]. Pick a point $(x,y)$ whose distances $d_1$ and $d_2$ from the focus $(ea,0)$ and directrix $x=\frac{a}{e}$, respectively, satisfy the condition for a hyperbola: $\frac{d_1}{d_2}=e>1$. Then

\[\begin{aligned} d_1^2 ~&=~ e^2 d_2^2\\ (x-ea)^2 ~+~ (y-0)^2 ~&=~ e^2\,\left(\left(x-\frac{a}{e}\right)^2 ~+~ (y-y)^2\right)\\ x^2 ~-~ \cancel{2eax} ~+~ e^2a^2 ~+~ y^2 ~&=~ e^2x^2 ~-~ \cancel{2eax} ~+~ a^2\

\[4pt] (e^2 - 1)\,x^2 ~-~ y^2 ~&=~ (e^2 - 1)\,a^2\\ %\frac{x^2}{a^2} ~-~ \frac{y^2}{(e^2 - 1)\,a^2} ~&=~ 1\

$$4pt] \frac{x^2}{a^2} ~-~ \frac{y^2}{b^2} ~&=~ 1\end{aligned} \nonumber$$

where $b^2 = (e^2 - 1)\,a^2 > 0$. The hyperbola thus has two branches ($x=\pm \frac{a}{b}\sqrt{y^2 + b^2}$), as in Figure [fig:hyperparts]. Let $c=ea$, so that $c>a$ and $b^2 = c^2-a^2$. By symmetry the hyperbola has two foci $(\pm c,0)$ and two directrices $x=\pm\frac{a^2}{c}$, and the lines $y=\pm \frac{b}{a}x$ are asymptotes. The vertexes are the points on the hyperbola closest to the directrices.

The center is the point midway between the foci. The transverse axis and conjugate axis are the perpendicular lines through the foci and center, respectively. In Figure [fig:hyperparts] the vertexes are $(\pm a,0)$, the $x$-axis is the transverse axis, the center is the origin $(0,0)$, and the conjugate axis is the $y$-axis. Note that the existence of two foci and directrices—when the definition of the hyperbola mentioned only a focus and a directrix—is simply a consequence of the symmetry about both axes imposed by the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. A parabola, in comparison, is symmetric about only one axis.

To see why the lines $y=\pm \frac{b}{a}x$ are asymptotes, consider the upper half $y=\frac{b}{a}\sqrt{x^2 - a^2}$ of both branches of the hyperbola. The difference between the line $y=\frac{b}{a}x$ and the upper right branch approaches zero as $x$ approaches infinity:

\[\begin{aligned} \lim_{x \to \infty} \,\left(\frac{b}{a}x ~-~ \frac{b}{a}\sqrt{x^2 - a^2}\right) ~&=~ \frac{b}{a}\,\lim_{x \to \infty} \,\left(x ~-~ \sqrt{x^2 - a^2}\,\right) \,\cdot\, \frac{x ~+~ \sqrt{x^2 - a^2}}{x ~+~ \sqrt{x^2 - a^2}}\

\[4pt] &=~ \frac{b}{a}\,\lim_{x \to \infty} \,\frac{x^2 ~-~ (x^2 - a^2)}{x ~+~ \sqrt{x^2 - a^2}}\

$$4pt] &=~ \lim_{x \to \infty} \,\frac{ab}{x ~+~ \sqrt{x^2 - a^2}} ~=~ 0\end{aligned} \nonumber$$

Thus the line $y=\frac{b}{a}x$ is an (oblique) asymptote for the upper half $y=\frac{b}{a}\sqrt{x^2 - a^2}$ of the right branch in the first quadrant of the $xy$-plane. So by symmetry the lines $y=\pm \frac{b}{a}x$ are asymptotes for both branches, i.e. for the entire hyperbola.

Conversely, given a hyperbola in the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, let $c^2=a^2+b^2$ to get the foci $(\pm c,0)$ and the directrices $x=\pm\frac{a^2}{c}$, while the eccentricity is $e = \frac{c}{a}$.

For the hyperbola $x^2 - y^2 = 1$ find the vertexes, foci, directrices, asymptotes and eccentricity.

Solution: Here $a=b=1$, so that $c^2=a^2+b^2=2$, i.e. $c=\sqrt{2}$. The vertexes are thus $(\pm 1,0)$, the asymptotes are $y=\pm x$, the foci are $(\pm\sqrt{2},0)$, the directrices are $x=\pm\frac{1}{\sqrt{2}}$, and the eccentricity is $e=\sqrt{2}$.

Switching the roles of $x$ and $y$ yields the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, shown in Figure [fig:hyperyx]. The transverse axis is the $y$-axis, the conjugate axis is the $x$-axis, the vertexes are at $(0,\pm a)$, and the foci are at $(0,\pm c)$, where $c^2 = a^2 + b^2$. The directrices are $y=\pm\frac{a^2}{c}$, the asymptotes are $y=\pm\frac{a}{b}x$, and the eccentricity is $e=\frac{c}{a}$.

For example, the hyperbola $y^2 - x^2 = 1$ has $a=b=1$, so that $c=\sqrt{2}$. The vertexes are $(0,\pm 1)$, the asymptotes are $y=\pm x$, the foci are $(0,\pm\sqrt{2})$, the directrices are $y=\pm\frac{1}{\sqrt{2}}$, and the eccentricity is $e=\sqrt{2}$. The hyperbola $y^2 - x^2 = 1$ is just the hyperbola $x^2 - y^2 = 1$ rotated $90\Degrees$. There is another way to define a hyperbola, in terms of two foci:

Figure [fig:hyperaltdefn] illustrates the above definition with foci $F_1$ and $F_2$. The difference $d_1-d_2$ of the distances $d_1$ and $d_2$ can be positive or negative depending on which branch the point $(x,y)$ is on, which is why the absolute value $\abs{d_1-d_2}$ is used.

It is left as an exercise to show that this second definition yields the same equation of the hyperbola as from the first definition. The second definition is often used as the primary definition in many textbooks, perhaps because it provides a simple way to construct a hyperbola by hand. Figure [fig:drawhyper] shows the procedure for foci $F_1$ and $F_2$: at the focus $F_1$ fasten one end of a ruler of length $L$, and at the other end $A$ of the ruler fasten one end of a string of length $L-d$ for some number $0<d<F_1F_2$. Fasten the other end of the string to the focus $F_2$ and hold the string taut with a pencil against the ruler at a point $P$, while rotating the ruler about $F_1$. The drawn figure will be one branch of a hyperbola, since the difference $PF_1 - PF_2$ will always be the positive constant $d$:

$$PF_1 - PF_2 ~=~ (L - AP) ~-~ PF_2 ~=~ L ~-~ (AP + PF_2) ~=~ L ~-~ (L - d) ~=~ d \nonumber$$

Reverse the roles of $F_1$ and $F_2$ to draw the other branch of the hyperbola.

By Exercise [exer:hyptan] in Section 3.4, the tangent line to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at a point $(x_0,y_0)$ is

$$\label{eqn:hypertan} \frac{x x_0}{a^2} ~-~ \frac{y y_0}{b^2} = 1 ~,$$

so that its slope is $\frac{b^2x_0}{a^2y_0}$ when $y_0 \ne 0$. Note that by the above equation, when $y_0=0$ (so that $x_0=\pm a$) the two tangent lines are the vertical lines $x=\pm a$. That slope will be used in proving the reflection property for the hyperbola: Light shone from one focus will reflect off the hyperbola in the opposite direction from the other focus. Figure [fig:hyperreflect] shows the light’s path from focus $F_2$ as it reflects at the point $P$ along the line through $P$ and the other focus $F_1$.

By Fermat’s Principle for curved surfaces, the reflection property is equivalent to saying the tangent line $L$ to the hyperbola at the point $P=(x_0,y_0)$ bisects the angle $\angle F_1PF_2$, i.e. $\theta_1=\theta_2$ as in Figure [fig:hyperreflpt1]. The reflection property holds trivially when $y_0=0$ (the light reflects straight back along the $x$-axis), so to show that $\theta_1=\theta_2$ assume $y_0 \ne 0$. By symmetry only $x_0 > 0$ need be considered. For the case $x_0 \ne c$, let $A$ be the $x$-intercept of the tangent line $L$. Then by Figure [fig:hyperreflpt1], since the sum of the angles in the triangle $\triangle F_1PA$ equals $180\Degrees$,

$$\alpha_1 + \theta_2 + (180\Degrees -\theta) = 180\Degrees \quad\Rightarrow\quad \theta_2 = \theta - \alpha_1 \quad\Rightarrow\quad \tan\,\theta_2 ~=~ \tan\,(\theta - \alpha_1) ~. \nonumber$$

Thus, since $\tan\,\theta$ is the slope of the tangent line $L$ (i.e. $\frac{b^2x_0}{a^2y_0}$), and $\tan\,\alpha_1 = \frac{y_0}{x_0 + c}$, then by the subtraction formula for the tangent function

\[\begin{aligned} \tan\,\theta_2 ~&=~ \frac{\tan\,\theta ~-~ \tan\,\alpha_1}{1 ~+~ \tan\,\theta\;\tan\,\alpha_1} ~=~ \frac{\frac{b^2x_0}{a^2y_0} ~-~ \frac{y_0}{x_0 ~+~ c}}{1 ~+~ \frac{b^2x_0}{a^2y_0}\;\cdot\;\frac{y_0}{x_0 ~+~ c}} ~=~ \frac{\frac{b^2 x_0^2 ~-~ a^2 y_0^2 ~+~ b^2 x_0 c}{\cancel{a^2 y_0\,(x_0 ~+~ c)}}}{\frac{(a^2 ~+~ b^2)\,x_0 y_0 ~+~ a^2 y_0 c}{\cancel{a^2 y_0\,(x_0 ~+~ c)}}}\

\[6pt] &=~ \frac{a^2 b^2 + b^2 x_0 c}{c^2 x_0 y_0 + a^2 y_0 c} \quad\text{(since $c^2 = a^2 + b^2$, and $\tfrac{x_0^2}{a^2} - \tfrac{y_0^2}{b^2} = 1 ~\Rightarrow~ b^2 x_0^2 ~-~ a^2 y_0^2 = a^2 b^2$)}\

$$6pt] &=~ \frac{b^2 \,\cancel{(a^2 ~+~ x_0 c)}}{cy_0\,\cancel{(a^2 ~+~ x_0 c)}} ~=~ \frac{b^2}{cy_0}\end{aligned} \nonumber$$

Similarly, since the sum of the angles in the triangle $\triangle F_2PA$ equals $180\Degrees$,

$$\alpha_2 + \theta_1 + \theta = 180\Degrees \quad\Rightarrow\quad \theta_1 = 180\Degrees - (\theta + \alpha_2) \quad\Rightarrow\quad \tan\,\theta_1 ~=~ -\tan\,(\theta + \alpha_2) ~. \nonumber$$

Thus, since $\tan\,\alpha_2 = -\tan\,(180\Degrees - \alpha_2) = -\frac{y_0}{x-c}$,

\[\begin{aligned} \tan\,\theta_1 ~&=~ -\frac{\tan\,\theta ~+~ \tan\,\alpha_2}{1 ~-~ \tan\,\theta\;\tan\,\alpha_2} ~=~ -\frac{\frac{b^2x_0}{a^2y_0} ~+~ \frac{-y_0}{x_0 ~-~ c}}{1 ~-~ \frac{b^2x_0}{a^2y_0}\;\cdot\;\frac{-y_0}{x_0 ~-~ c}} ~=~ -\frac{\frac{b^2 x_0^2 ~-~ a^2 y_0^2 ~-~ b^2 x_0 c}{\cancel{a^2 y_0\,(x_0 ~-~ c)}}}{\frac{(a^2 ~+~ b^2)\,x_0 y_0 ~-~ a^2 y_0 c}{\cancel{a^2 y_0\,(x_0 ~-~ c)}}}\

$$6pt] &=~ -\frac{a^2 b^2 - b^2 x_0 c}{c^2 x_0 y_0 - a^2 y_0 c} ~=~ \frac{b^2 \,\cancel{(a^2 ~-~ x_0 c)}}{cy_0\,\cancel{(a^2 ~-~ x_0 c)}} ~=~ \frac{b^2}{cy_0} ~=~ \tan\,\theta_2 ~,\end{aligned} \nonumber$$

i.e. $\theta_1 = \theta_2$ (since $0\Degrees < \theta_1,\,\theta_2 < 90\Degrees$). $\checkmark\quad$ Note: The case $x_0=c$ is left as an exercise.

Ellipses, parabolas and hyperbolas are sometimes called conic sections, due to being formed by intersections of planes with a double circular cone of unlimited extent:

Each double cone in Figure [fig:conics] has two nappes—a cone extending upward and one extending downward. When a plane intersects only one nappe in a closed noncircular curve, as in Figure [fig:conics](a), that curve is an ellipse. A plane that is parallel to a line on one nappe, as in Figure [fig:conics](b), intersects only that nappe in a parabola. The intersection of a plane with both nappes, as in Figure [fig:conics](c), is a hyperbola.

To prove that the ellipse, parabola and hyperbola really are represented by the indicated conic sections, first a minor result is needed from three-dimensional geometry: tangent line segments to a sphere from the same point have equal lengths, as in Figure [fig:spheretan]. Since the right triangles $\triangle QOP_1$ and $\triangle QOP_2$ share the same hypotenuse $\overline{QO}$ and have legs $\overline{OP_1}$ and $\overline{OP_2}$ of equal length (the radius of the sphere), the result $QP_1=QP_2$ follows by the Pythagorean Theorem. For the case of a right circular double cone (i.e. the base of each nappe is a circle in a plane perpendicular to the axis of the cone⁷), let $\beta$ be the complement of the angle that the cone makes with its axis, as in Figure [fig:conicproof]. So $\beta$ is the angle the cone makes with any circular base of the cone. This constant angle $\beta$, with $0\Degrees < \beta < 90\Degrees$, is an intrinsic property of the cone. Let $P_c$ be a plane that intersects the lower nappe of the cone in a curve $C$, such that $P_c$ makes an angle $\alpha$ with any base circle of the cone. By symmetry, only the angles $0\Degrees < \alpha \le 90\Degrees$ need be considered.⁸ Inscribe a sphere in the cone so that it touches $P_c$ at a point $F$, as in Figure [fig:conicproof] ($P_c$ is the tangent plane to the sphere at $F$). Let $P_0$ be the plane through the circle where the inscribed sphere touches the cone, and let $D$ be the line of intersection of the planes $P_c$ and $P_0$. It will be shown that $F$ and $D$ are the focus and directrix, respectively, of the curve $C$.

Let $P$ be any point on the curve $C$, then let $Q$ be the point on the plane $P_0$ that lies on a line through $P$ and the cone’s vertex. Drop a perpendicular line segment from $P$ to the point $A$ in the plane $P_0$. From $A$ draw a perpendicular line segment to the point $G$ on the line $D$. Then as Figure [fig:conicproof] shows, $\triangle QAP$ and $\triangle PAG$ are right triangles, with

$$\sin\,\alpha ~=~ \frac{PA}{PG} \quad\text{and}\quad \sin\,\beta ~=~ \frac{PA}{PQ} ~. \nonumber$$

However, since $\overline{PF}$ and $\overline{PQ}$ are both tangent line segments to the inscribed sphere from the same point $P$, the result proved earlier shows that $PQ = PF$. Thus,

$$PG\,\sin\,\alpha ~=~ PA ~=~ PQ\,\sin\,\beta ~=~ PF\,\sin\,\beta \quad\Rightarrow\quad \frac{PF}{PG} ~=~ \frac{\sin\,\alpha}{\sin\,\beta} ~. \nonumber$$

Let $e = \frac{PF}{PG}$. Then $e$ is the same constant $\frac{\sin\,\alpha}{\sin\,\beta}$ for any point $P$ on the curve $C$. Thus, by definition, $e$ is the eccentricity of the curve $C$ with focus $F$ and directrix $D$. If $0\Degrees < \alpha < \beta$ then $0\Degrees < \sin\,\alpha < \sin\,\beta$ so that $0 < \frac{\sin\,\alpha}{\sin\,\beta} < 1$, which means that $0 < e < 1$, and hence $C$ is an ellipse (by the second definition of an ellipse). Likewise, if $\alpha = \beta$ then $e=1$, so that $C$ is a parabola. Finally, if $\beta < \alpha \le 90\Degrees$ then $e > 1$, so that $C$ is a hyperbola (and will intersect both nappes of the cone). Thus, the ellipse, parabola and hyperbola truly are conic sections. $\checkmark$ [sec7dot3]

Construct a hyperbola using the procedure shown in Figure [fig:drawhyper]. Place the two focus pins 7in apart and use a 9in piece of string attached to a 12in ruler.

For Exercises 2-6, sketch the graph of the given hyperbola, indicate the exact locations of the foci and vertexes, indicate each directrix and asymptote, and find the eccentricity $e$. [[1.]]

5

$\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$

$\dfrac{x^2}{8} - \dfrac{y^2}{15} = 1$

$\dfrac{4x^2}{25} - \dfrac{y^2}{4} = 1$

$x^2 - 4y^2 = 1\vphantom{\dfrac{x^2}{16}}$

$25y^2 - 9x^2 = 225\vphantom{\dfrac{x^2}{16}}$

Find the equation of the hyperbola with foci $(\pm 5,0)$ and vertexes $(\pm 3,0)$. [[1.]]

In the second definition of the hyperbola on p.219, let $2a$ be the constant absolute value of the difference of distances from points on the hyperbola to the foci, for some $a>0$. Let the foci be at $(\pm c,0)$ for some $c>0$. Show that this second definition yields a hyperbola having an equation of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2}=1$, with $b > 0$.

Prove the reflection property for the hyperbola (see pp.220-221) when $x_0=c$ for the point $P=(x_0,y_0)$ on the hyperbola with foci at $(\pm c,0)$.

Show that a straight line parallel to an asymptote of a hyperbola intersects the hyperbola at exactly one point.

A latus rectum (plural: latus recta) of a hyperbola is a chord through either focus perpendicular to the transverse axis. Show that the latus recta of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2}=1$ have length $\frac{2b^2}{a}$.

Show that the segment of an asymptote of a hyperbola between the two directrices has the same length as the line segment between the vertexes.

Show that the segment of a tangent line to a hyperbola between the hyperbola’s asymptotes has its midpoint at the point of tangency.

The focal radii of a hyperbola are the line segments from the foci to points on the hyperbola. Show that the lengths of the focal radii to points $(x,y)$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2}=1$ are $a+ex$ and $ex-a$, where $e$ is the eccentricity.

Show that a tangent line to a hyperbola together with the hyperbola’s asymptotes bounds a triangle of constant area (i.e. the area is independent of the point of tangency on the hyperbola).

Show that the product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2}=1$ to the asymptotes $y = \pm\frac{b}{a}x$ is a constant.

A person at a point $P=(x,y)$ hears the crack of a rifle located at the point $F_1=(-1000,0)$ and the sound of the fired bullet hitting its target located at the point $F_2=(1000,0)$ at the same time. The bullet’s speed is 2000 ft/sec and the speed of sound is 1100 ft/sec. Find an equation relating $x$ and $y$.

Show that the set of all midpoints of a family of parallel chords either in one branch or between the two branches of a hyperbola lie on a line through the center of the hyperbola.

Prove a second reflection property of hyperbolas: a light shone between the two branches and directed toward one focus will reflect toward the other focus.