for all values of the parameter t. The angles αand βconstructed at the point P can
be calculated from the dot products
(−ˆet)·ˆer 2 = cos α=asinh t(acosh t+c) + b^2 sinh tcosh t
(√
a^2 sinh^2 t+b^2 cosh^2 t)(√
(acosh t+c)^2 +b^2 sinh^2 t)=asinh t(acosh t+c) + b^2 sinh tcosh t
(√
a^2 sinh^2 t+b^2 cosh^2 t)(√
(acosh t−c)^2 +b^2 sinh^2 t+ 2 a)(7 .8)(−ˆet)·ˆer 1 = cos β=asinh t(acosh t−c) + b^2 sinh tcosh t
(√
a^2 sinh^2 t+b^2 cosh^2 t)(√
(acosh t−c)^2 +b^2 sinh^2 t)(7 .9)If cos α= cosβ then one must show that the right-hand sides of equations (7.8)
and (7.9) are equal. Setting the right-hand sides equal to one another and simplifying
produces
a^2 sinh√tcosh t+ac sinh t+b^2 sinh tcosh t
(acosh t−c)^2 +b^2 sinh^2 t+ 2 a=a^2 sinh √tcosh t−ac sinh t+b^2 sinh tcosh t
(acosh t−c)^2 +b^2 sinh^2 t(7 .10)To show equation (7.10) reduces to an identity, first show equation (7.10) can be
written
|r 2 |=ccosh t−a
ccosh t+a√
(acosh t−c)^2 +b^2 sinh^2 t+ 2 aand then use the fact that c^2 =a^2 +b^2 and |r 2 |=ccosh t−ato simplify the above
equation to the form
ccosh t+a=√
(acosh t−c)^2 +b^2 sinh^2 t+ 2 a (7 .11)It is now an easy exercise to show equation (7.11) reduces to an identity.
All this algebra shows that the angles αand βare equal and consequently the
complementary angles θ 1 and θ 2 are also equal, showing the hyperbola has the prop-
erty that the angle of incidence equals the angle of reflection. The above results
imply that a ray of light aimed at the focus F 2 will be reflected and pass through the
other focus. This reflection property of the hyperbola is one of the basic principles
used in the construction of a reflecting telescope.