As locus of points A hyperbola can be defined geometrically as a
set of points (
locus of points) in the Euclidean plane: {{block indent |em=1.5 |text= A
hyperbola is a set of points, such that for any point P of the set, the absolute difference of the distances |PF_1|,\, |PF_2| to two fixed points F_1, F_2 (the
foci) is constant, usually denoted by H = \left\{P : \left|\left|PF_2\right| - \left|PF_1\right|\right| = 2a \right\} . }} The midpoint M of the line segment joining the foci is called the
center of the hyperbola. The line through the foci is called the
major axis. It contains the
vertices V_1, V_2, which have distance a to the center. The distance c of the foci to the center is called the
focal distance or
linear eccentricity. The quotient \tfrac c a is the
eccentricity e. The equation \left|\left|PF_2\right| - \left|PF_1\right|\right| = 2a can be viewed in a different way (see diagram): If c_2 is the circle with midpoint F_2 and radius 2a, then the distance of a point P of the right branch to the circle c_2 equals the distance to the focus F_1: |PF_1|=|Pc_2|. c_2 is called the
circular directrix (related to focus F_2) of the hyperbola. In order to get the left branch of the hyperbola, one has to use the circular directrix related to F_1. This property should not be confused with the definition of a hyperbola with help of a directrix (line) below.
Hyperbola with equation If the
xy-coordinate system is
rotated about the origin by the angle +45^\circ and new coordinates \xi,\eta are assigned, then x = \tfrac{\xi+\eta}{\sqrt{2}},\; y = \tfrac{-\xi+\eta}{\sqrt{2}} . The rectangular hyperbola \tfrac{x^2-y^2}{a^2} = 1 (whose
semi-axes are equal) has the new equation \tfrac{2\xi\eta}{a^2} = 1. Solving for \eta yields \eta = \tfrac{a^2/2}{\xi} \ . Thus, in an
xy-coordinate system the
graph of a function f: x \mapsto \tfrac{A}{x},\; A>0\; , with equation y = \frac{A}{x}\;, A>0\; , is a
rectangular hyperbola entirely in the first and third
quadrants with • the coordinate axes as
asymptotes, • the line y = x as
major axis , • the
center (0,0) and the
semi-axis a = b = \sqrt{2A} \; , • the
vertices \left(\sqrt{A},\sqrt{A}\right), \left(-\sqrt{A},-\sqrt{A}\right) \; , • the
semi-latus rectum and
radius of curvature at the vertices p=a=\sqrt{2A} \; , • the
linear eccentricity c=2\sqrt{A} and the eccentricity e=\sqrt{2} \; , • the
tangent y=-\tfrac{A}{x_0^2}x+2\tfrac{A}{x_0} at point (x_0,A/x_0)\; . A rotation of the original hyperbola by -45^\circ results in a rectangular hyperbola entirely in the second and fourth quadrants, with the same asymptotes, center, semi-latus rectum, radius of curvature at the vertices, linear eccentricity, and eccentricity as for the case of +45^\circ rotation, with equation y = -\frac{A}{x} \; , ~~ A>0\; , • the
semi-axes a = b = \sqrt{2A} \; , • the line y = -x as major axis, • the
vertices \left(-\sqrt{A},\sqrt{A}\right), \left(\sqrt{A},-\sqrt{A}\right) \; . Shifting the hyperbola with equation y=\frac{A}{x}, \ A\ne 0\ , so that the new center is yields the new equation y=\frac{A}{x-c_0}+d_0\; , and the new asymptotes are x=c_0 and y=d_0. The shape parameters a,b,p,c,e remain unchanged.
By the directrix property The two lines at distance d = \frac{a^2}c from the center and parallel to the minor axis are called
directrices of the hyperbola (see diagram). For an arbitrary point P of the hyperbola the quotient of the distance to one focus and to the corresponding directrix (see diagram) is equal to the eccentricity: \frac = \frac = e= \frac{c}{a} \, . The proof for the pair F_1, l_1 follows from the fact that |PF_1|^2 = (x-c)^2+y^2,\ |Pl_1|^2 = \left(x-\tfrac{a^2}{c}\right)^2 and y^2 = \tfrac{b^2}{a^2}x^2-b^2 satisfy the equation |PF_1|^2-\frac{c^2}{a^2}|Pl_1|^2 = 0\ . The second case is proven analogously. The
inverse statement is also true and can be used to define a hyperbola (in a manner similar to the definition of a parabola): For any point F (focus), any line l (directrix) not through F and any
real number e with e > 1 the set of points (locus of points), for which the quotient of the distances to the point and to the line is e H = \left\{P \, \Biggr| \, \frac = e\right\} is a hyperbola. (The choice e = 1 yields a
parabola and if e an
ellipse.)
Proof Let F=(f,0) ,\ e >0 and assume (0,0) is a point on the curve. The directrix l has equation x=-\tfrac{f}{e}. With P=(x,y), the relation |PF|^2 = e^2|Pl|^2 produces the equations :(x-f)^2+y^2 = e^2\left(x+\tfrac{f}{e}\right)^2 = (e x+f)^2 and x^2(e^2-1)+2xf(1+e)-y^2 = 0. The substitution p=f(1+e) yields x^2(e^2-1)+2px-y^2 = 0. This is the equation of an
ellipse (e) or a
parabola (e=1) or a
hyperbola (e>1). All of these non-degenerate conics have, in common, the origin as a vertex (see diagram). If e > 1, introduce new parameters a,b so that e^2-1 = \tfrac{b^2}{a^2}, \text { and }\ p = \tfrac{b^2}{a}, and then the equation above becomes \frac{(x+a)^2}{a^2} - \frac{y^2}{b^2} = 1 \, , which is the equation of a hyperbola with center (-a,0), the
x-axis as major axis and the major/minor semi axis a,b.
Construction of a directrix Because of c \cdot \tfrac{a^2}{c}=a^2 point L_1 of directrix l_1 (see diagram) and focus F_1 are inverse with respect to the
circle inversion at circle x^2+y^2=a^2 (in diagram green). Hence point E_1 can be constructed using the
theorem of Thales (not shown in the diagram). The directrix l_1 is the perpendicular to line \overline{F_1F_2} through point E_1.
Alternative construction of E_1: Calculation shows, that point E_1 is the intersection of the asymptote with its perpendicular through F_1 (see diagram).
As plane section of a cone The intersection of an upright double cone by a plane not through the vertex with slope greater than the slope of the lines on the cone is a hyperbola (see diagram: red curve). In order to prove the defining property of a hyperbola (see above) one uses two
Dandelin spheres d_1, d_2, which are spheres that touch the cone along circles c_2 and the intersecting (hyperbola) plane at points F_1 and It turns out: F_1, F_2 are the
foci of the hyperbola. • Let P be an arbitrary point of the intersection curve. • The
generatrix of the cone containing P intersects circle c_1 at point A and circle c_2 at a point B. • The line segments \overline{PF_1} and \overline{PA} are tangential to the sphere d_1 and, hence, are of equal length. • The line segments \overline{PF_2} and \overline{PB} are tangential to the sphere d_2 and, hence, are of equal length. • The result is: |PF_1| - |PF_2| = |PA| - |PB| = |AB| is independent of the hyperbola point because no matter where point P is, A, B have to be on circles and line segment AB has to cross the apex. Therefore, as point P moves along the red curve (hyperbola), line segment \overline{AB} simply rotates about apex without changing its length.
Pin and string construction The definition of a hyperbola by its foci and its circular directrices (see above) can be used for drawing an arc of it with help of pins, a string and a ruler: • Choose the
foci F_1,F_2 and one of the
circular directrices, for example c_2 (circle with radius 2a) • A
ruler is fixed at point F_2 free to rotate around F_2. Point B is marked at distance 2a. • A
string gets its one end pinned at point A on the ruler and its length is made |AB|. • The free end of the string is pinned to point F_1. • Take a
pen and hold the string tight to the edge of the ruler. •
Rotating the ruler around F_2 prompts the pen to draw an arc of the right branch of the hyperbola, because of |PF_1| = |PB| (see the definition of a hyperbola by
circular directrices).
Steiner generation of a hyperbola The following method to construct single points of a hyperbola relies on the
Steiner generation of a non degenerate conic section: For the generation of points of the hyperbola \tfrac{x^2}{a^2}-\tfrac{y^2}{b^2} = 1 one uses the pencils at the vertices V_1,V_2. Let P = (x_0,y_0) be a point of the hyperbola and A = (a,y_0), B = (x_0,0). The line segment \overline{BP} is divided into n equally-spaced segments and this division is projected parallel with the diagonal AB as direction onto the line segment \overline{AP} (see diagram). The
parallel projection is part of the projective mapping between the pencils at V_1 and V_2 needed. The intersection points of any two related lines S_1 A_i and S_2 B_i are points of the uniquely defined hyperbola.
Remarks: • The subdivision could be extended beyond the points A and B in order to get more points, but the determination of the intersection points would become more inaccurate. A better idea is extending the points already constructed by symmetry (see animation). • The Steiner generation exists for ellipses and parabolas, too. • The Steiner generation is sometimes called a
parallelogram method because one can use other points rather than the vertices, which starts with a
parallelogram instead of a rectangle.
Inscribed angles for hyperbolas and the 3-point-form A hyperbola with equation y=\tfrac{a}{x-b}+c,\ a \ne 0 is uniquely determined by three points (x_1,y_1),\;(x_2,y_2),\; (x_3,y_3) with different
x- and
y-coordinates. A simple way to determine the shape parameters a,b,c uses the
inscribed angle theorem for hyperbolas: {{block indent |em=1.5 |text=In order to
measure an angle between two lines with equations y=m_1x+d_1, \ y=m_2x + d_2\ ,m_1,m_2 \ne 0 in this context one uses the quotient \frac{m_1}{m_2}\ .}} Analogous to the
inscribed angle theorem for circles one gets the {{math theorem The four points are on a hyperbola with equation y = \tfrac{a}{x-b} + c if and only if the angles at P_3 and P_4 are equal in the sense of the measurement above. That means if \frac{(y_4-y_1)}{(x_4-x_1)}\frac{(x_4-x_2)}{(y_4-y_2)}=\frac{(y_3-y_1)}{(x_3-x_1)}\frac{(x_3-x_2)}{(y_3-y_2)} The proof can be derived by straightforward calculation. If the points are on a hyperbola, one can assume the hyperbola's equation is }} A consequence of the inscribed angle theorem for hyperbolas is the
As an affine image of the unit hyperbola Another definition of a hyperbola uses
affine transformations:
Parametric representation An affine transformation of the Euclidean plane has the form \vec x \to \vec f_0+A\vec x, where A is a regular
matrix (its
determinant is not 0) and \vec f_0 is an arbitrary vector. If \vec f_1, \vec f_2 are the column vectors of the matrix A, the unit hyperbola (\pm\cosh(t),\sinh(t)), t \in \R, is mapped onto the hyperbola \vec x = \vec p(t)=\vec f_0 \pm\vec f_1 \cosh t +\vec f_2 \sinh t \ . \vec f_0 is the center, \vec f_0+ \vec f_1 a point of the hyperbola and \vec f_2 a tangent vector at this point.
Vertices In general the vectors \vec f_1, \vec f_2 are not perpendicular. That means, in general \vec f_0\pm \vec f_1 are
not the vertices of the hyperbola. But \vec f_1\pm \vec f_2 point into the directions of the asymptotes. The tangent vector at point \vec p(t) is \vec p'(t) = \vec f_1\sinh t + \vec f_2\cosh t \ . Because at a vertex the tangent is perpendicular to the major axis of the hyperbola one gets the parameter t_0 of a vertex from the equation \vec p'(t)\cdot \left(\vec p(t) -\vec f_0\right) = \left(\vec f_1\sinh t + \vec f_2\cosh t\right) \cdot \left(\vec f_1 \cosh t +\vec f_2 \sinh t\right) = 0 and hence from \coth (2t_0)= -\tfrac{\vec f_1^{\, 2}+\vec f_2^{\, 2}}{2\vec f_1 \cdot \vec f_2} \ , which yields t_0=\tfrac{1}{4}\ln\tfrac{\left(\vec f_1-\vec f_2\right)^2}{\left(\vec f_1+\vec f_2\right)^2}. The formulae and \operatorname{arcoth} x = \tfrac{1}{2}\ln\tfrac{x+1}{x-1} were used. The two
vertices of the hyperbola are \vec f_0\pm\left(\vec f_1\cosh t_0 +\vec f_2 \sinh t_0\right).
Implicit representation Solving the parametric representation for \cosh t, \sinh t by
Cramer's rule and using \;\cosh^2t-\sinh^2t -1 = 0\; , one gets the implicit representation \det\left(\vec x\!-\!\vec f\!_0,\vec f\!_2\right)^2 - \det\left(\vec f\!_1,\vec x\!-\!\vec f\!_0\right)^2 - \det\left(\vec f\!_1,\vec f\!_2\right)^2 = 0 .
Hyperbola in space The definition of a hyperbola in this section gives a parametric representation of an arbitrary hyperbola, even in space, if one allows \vec f\!_0, \vec f\!_1, \vec f\!_2 to be vectors in space.
As an affine image of the hyperbola Because the unit hyperbola x^2-y^2=1 is affinely equivalent to the hyperbola y=1/x, an arbitrary hyperbola can be considered as the affine image (see previous section) of the hyperbola \vec x = \vec p(t) = \vec f_0 + \vec f_1 t + \vec f_2 \tfrac{1}{t}, \quad t\ne 0\, . M: \vec f_0 is the center of the hyperbola, the vectors \vec f_1 , \vec f_2 have the directions of the asymptotes and \vec f_1 + \vec f_2 is a point of the hyperbola. The tangent vector is \vec p'(t)=\vec f_1 - \vec f_2 \tfrac{1}{t^2}. At a vertex the tangent is perpendicular to the major axis. Hence \vec p'(t)\cdot \left(\vec p(t) -\vec f_0\right) = \left(\vec f_1 - \vec f_2 \tfrac{1}{t^2}\right)\cdot\left(\vec f_1 t+ \vec f_2 \tfrac{1}{t}\right) = \vec f_1^2t-\vec f_2^2 \tfrac{1}{t^3} = 0 and the parameter of a vertex is t_0= \pm \sqrt[4]{\frac{\vec f_2^2}{\vec f_1^2}}. \left|\vec f\!_1\right| = \left|\vec f\!_2\right| is equivalent to t_0 = \pm 1 and \vec f_0 \pm (\vec f_1+\vec f_2) are the vertices of the hyperbola. The following properties of a hyperbola are easily proven using the representation of a hyperbola introduced in this section.
Tangent construction The tangent vector can be rewritten by factorization: \vec p'(t)=\tfrac{1}{t}\left(\vec f_1t - \vec f_2 \tfrac{1}{t}\right) \ . This means that {{block indent |em=1.5 |text=the diagonal AB of the parallelogram M: \ \vec f_0, \ A=\vec f_0+\vec f_1t,\ B:\ \vec f_0+ \vec f_2 \tfrac{1}{t},\ P:\ \vec f_0+\vec f_1t+\vec f_2 \tfrac{1}{t} is parallel to the tangent at the hyperbola point P (see diagram).}} This property provides a way to construct the tangent at a point on the hyperbola. This property of a hyperbola is an affine version of the 3-point-degeneration of
Pascal's theorem. ;Area of the grey parallelogram: The area of the grey parallelogram MAPB in the above diagram is \text{Area} = \left|\det\left( t\vec f_1, \tfrac{1}{t}\vec f_2\right)\right| = \left|\det\left(\vec f_1,\vec f_2\right)\right| = \cdots = \frac{ab}{2} and hence independent of point P. The last equation follows from a calculation for the case, where P is a vertex and the hyperbola in its canonical form \tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}=1 \, .
Point construction For a hyperbola with parametric representation \vec x = \vec p(t) = \vec f_1 t + \vec f_2 \tfrac{1}{t} (for simplicity the center is the origin) the following is true: {{block indent |em=1.5 |text=For any two points P_1:\ \vec f_1 t_1+ \vec f_2 \tfrac{1}{t_1},\ P_2:\ \vec f_1 t_2+ \vec f_2 \tfrac{1}{t_2} the points A:\ \vec a =\vec f_1 t_1+ \vec f_2 \tfrac{1}{t_2}, \ B:\ \vec b=\vec f_1 t_2+ \vec f_2 \tfrac{1}{t_1} are collinear with the center of the hyperbola (see diagram).}} The simple proof is a consequence of the equation \tfrac{1}{t_1}\vec a = \tfrac{1}{t_2}\vec b. This property provides a possibility to construct points of a hyperbola if the asymptotes and one point are given. This property of a hyperbola is an affine version of the 4-point-degeneration of
Pascal's theorem.
Tangent–asymptotes triangle For simplicity the center of the hyperbola may be the origin and the vectors \vec f_1,\vec f_2 have equal length. If the last assumption is not fulfilled one can first apply a parameter transformation (see above) in order to make the assumption true. Hence \pm (\vec f_1 + \vec f_2) are the vertices, \pm(\vec f_1-\vec f_2) span the minor axis and one gets |\vec f_1 + \vec f_2| = a and |\vec f_1 - \vec f_2| = b. For the intersection points of the tangent at point \vec p(t_0) = \vec f_1 t_0 + \vec f_2 \tfrac{1}{t_0} with the asymptotes one gets the points C = 2t_0\vec f_1,\ D = \tfrac{2}{t_0}\vec f_2. The
area of the triangle M,C,D can be calculated by a 2 × 2 determinant: A = \tfrac{1}{2}\Big|\det\left( 2t_0\vec f_1, \tfrac{2}{t_0}\vec f_2\right)\Big| = 2\Big|\det\left(\vec f_1,\vec f_2\right)\Big| (see rules for
determinants). \left|\det(\vec f_1,\vec f_2)\right| is the area of the rhombus generated by \vec f_1,\vec f_2. The area of a rhombus is equal to one half of the product of its diagonals. The diagonals are the semi-axes a,b of the hyperbola. Hence:
Reciprocation of a circle The
reciprocation of a
circle B in a circle
C always yields a conic section such as a hyperbola. The process of "reciprocation in a circle
C" consists of replacing every line and point in a geometrical figure with their corresponding
pole and polar, respectively. The
pole of a line is the
inversion of its closest point to the circle
C, whereas the polar of a point is the converse, namely, a line whose closest point to
C is the inversion of the point. The eccentricity of the conic section obtained by reciprocation is the ratio of the distances between the two circles' centers to the radius
r of reciprocation circle
C. If
B and
C represent the points at the centers of the corresponding circles, then e = \frac{\overline{BC}}{r}. Since the eccentricity of a hyperbola is always greater than one, the center
B must lie outside of the reciprocating circle
C. This definition implies that the hyperbola is both the
locus of the poles of the tangent lines to the circle
B, as well as the
envelope of the polar lines of the points on
B. Conversely, the circle
B is the envelope of polars of points on the hyperbola, and the locus of poles of tangent lines to the hyperbola. Two tangent lines to
B have no (finite) poles because they pass through the center
C of the reciprocation circle
C; the polars of the corresponding tangent points on
B are the asymptotes of the hyperbola. The two branches of the hyperbola correspond to the two parts of the circle
B that are separated by these tangent points.
Quadratic equation A hyperbola can also be defined as a second-degree equation in the Cartesian coordinates (x, y) in the
plane, A_{xx} x^2 + 2 A_{xy} xy + A_{yy} y^2 + 2 B_x x + 2 B_y y + C = 0, provided that the constants A_{xx}, A_{xy}, A_{yy}, B_x, B_y, and C satisfy the determinant condition D := \begin{vmatrix} A_{xx} & A_{xy} \\ A_{xy} & A_{yy} \end{vmatrix} This determinant is conventionally called the
discriminant of the conic section. A special case of a hyperbola—the
degenerate hyperbola consisting of two intersecting lines—occurs when another determinant is zero: \Delta := \begin{vmatrix} A_{xx} & A_{xy} & B_x \\ A_{xy} & A_{yy} & B_y \\ B_x & B_y & C \end{vmatrix} = 0. This determinant \Delta is sometimes called the discriminant of the conic section. The general equation's coefficients can be obtained from known semi-major axis a, semi-minor axis b, center coordinates (x_\circ, y_\circ), and rotation angle \theta (the angle from the positive horizontal axis to the hyperbola's major axis) using the formulae: \begin{align} A_{xx} &= -a^2 \sin^2\theta + b^2 \cos^2\theta, & B_{x} &= -A_{xx} x_\circ - A_{xy} y_\circ, \\[1ex] A_{yy} &= -a^2 \cos^2\theta + b^2 \sin^2\theta, & B_{y} &= - A_{xy} x_\circ - A_{yy} y_\circ, \\[1ex] A_{xy} &= \left(a^2 + b^2\right) \sin\theta \cos\theta, & C &= A_{xx} x_\circ^2 + 2A_{xy} x_\circ y_\circ + A_{yy} y_\circ^2 - a^2 b^2. \end{align} These expressions can be derived from the canonical equation \frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1 by a
translation and rotation of the coordinates \begin{alignat}{2} X &= \phantom+\left(x - x_\circ\right) \cos\theta &&+ \left(y - y_\circ\right) \sin\theta, \\ Y &= -\left(x - x_\circ\right) \sin\theta &&+ \left(y - y_\circ\right) \cos\theta. \end{alignat} Given the above general parametrization of the hyperbola in Cartesian coordinates, the eccentricity can be found using the formula in Conic section#Eccentricity in terms of coefficients. The center (x_c, y_c) of the hyperbola may be determined from the formulae \begin{align} x_c &= -\frac{1}{D} \, \begin{vmatrix} B_x & A_{xy} \\ B_y & A_{yy} \end{vmatrix} \,, \\[1ex] y_c &= -\frac{1}{D} \, \begin{vmatrix} A_{xx} & B_x \\ A_{xy} & B_y \end{vmatrix} \,. \end{align} In terms of new coordinates, \xi = x - x_c and \eta = y - y_c, the defining equation of the hyperbola can be written A_{xx} \xi^2 + 2A_{xy} \xi\eta + A_{yy} \eta^2 + \frac \Delta D = 0. The principal axes of the hyperbola make an angle \varphi with the positive x-axis that is given by \tan (2\varphi) = \frac{2A_{xy}}{A_{xx} - A_{yy}}. Rotating the coordinate axes so that the x-axis is aligned with the transverse axis brings the equation into its
canonical form \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. The major and minor semiaxes a and b are defined by the equations \begin{align} a^2 &= -\frac{\Delta}{\lambda_1 D} = -\frac{\Delta}{\lambda_1^2 \lambda_2}, \\[1ex] b^2 &= -\frac{\Delta}{\lambda_2 D} = -\frac{\Delta}{\lambda_1 \lambda_2^2}, \end{align} where \lambda_1 and \lambda_2 are the
roots of the
quadratic equation \lambda^2 - \left( A_{xx} + A_{yy} \right)\lambda + D = 0. For comparison, the corresponding equation for a degenerate hyperbola (consisting of two intersecting lines) is \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0. The tangent line to a given point (x_0, y_0) on the hyperbola is defined by the equation E x + F y + G = 0 where E, F, and G are defined by \begin{align} E &= A_{xx} x_0 + A_{xy} y_0 + B_x, \\[1ex] F &= A_{xy} x_0 + A_{yy} y_0 + B_y, \\[1ex] G &= B_x x_0 + B_y y_0 + C. \end{align} The
normal line to the hyperbola at the same point is given by the equation F(x - x_0) - E(y - y_0) = 0. The normal line is perpendicular to the tangent line, and both pass through the same point (x_0, y_0). From the equation \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \qquad 0 the left focus is (-ae,0) and the right focus is (ae,0), where e is the eccentricity. Denote the distances from a point (x, y) to the left and right foci as r_1 and r_2. For a point on the right branch, r_1 - r_2 = 2 a, and for a point on the left branch, r_2 - r_1 = 2 a. This can be proved as follows: If (x, y) is a point on the hyperbola the distance to the left focal point is r_1^2 = (x+a e)^2 + y^2 = x^2 + 2 x a e + a^2 e^2 + \left(x^2-a^2\right) \left(e^2-1\right) = (e x + a)^2. To the right focal point the distance is r_2^2 = (x-a e)^2 + y^2 = x^2 - 2 x a e + a^2 e^2 + \left(x^2-a^2\right) \left(e^2-1\right) = (e x - a)^2. If (x, y) is a point on the right branch of the hyperbola then ex > a and \begin{align} r_1 &= e x + a, \\ r_2 &= e x - a. \end{align} Subtracting these equations one gets r_1 - r_2 = 2a. If (x, y) is a point on the left branch of the hyperbola then ex and \begin{align} r_1 &= - e x - a, \\ r_2 &= - e x + a. \end{align} Subtracting these equations one gets r_2 - r_1 = 2a. ==In Cartesian coordinates==