complementary angles associated with these angles are labeled αand βrespectively.
Next construct the vector r 1 running from the point P on the hyperbola to the focus
F 1 and then construct the vector r 2 running from the P to the other focus F 2 as
illustrated in the figure 7-3.
The position vector to a general point on the right branch of the hyperbola can
be expressed
r =r (t) = acosh teˆ 1 +bsinh teˆ 2
where tis a parameter. The value of the parameter t, call it t 0 , corresponding to the
point P having coordinates (x 0 , y 0 )is obtained by solving the equations
x 0 =acosh t and y 0 =bsinh t
simultaneously to obtain t 0 = tanh −^1
(
ay 0
bx 0
)
. The derivative of the position vector is
dr
dt
=asinh tˆe 1 +bcosh tˆe 2
and this vector, when evaluated using the parameter value t 0 is a tangent vector to
the hyperbola at the point P. The vector
ˆet=∣∣dr^1
dt
∣∣dr
dt
=√asinh tˆe^1 +bcosh tˆe^2
a^2 sinh^2 t+b^2 cosh^2 t
also evaluated at t 0 , is a unit tangent vector to the hyperbola at the point P. Using
vector addition one can show that the vectors r 1 and r 2 are given by
r 1 =−(acosh t+c)ˆe 1 −bsinh t
r 2 =−(acosh t−c)ˆe 1 −bsinh tˆe 2
all to be evaluated at the parameter value t 0. Unit vectors in the directions of r 1
and r 2 are
ˆer 1 =−(acosh t+c)
ˆe 1 −bsinh tˆe 2
√
(acosh t+c)^2 +b^2 sinh^2 t
ˆer 2 =−√(acosh t−c)ˆe^1 −bsinh tˆe^2
(acosh t−c)^2 +b^2 sinh^2 t
also to be evaluated at t=t 0. The given hyperbola satisfies the properties that
a^2 +b^2 =c^2 and |r 1 |−|r 2 |= 2aor
√
(acosh t+c)^2 +b^2 sinh^2 t−
√
(acosh t−c)^2 +b^2 sinh^2 t= 2a
