Begin2.DVI

in figure 7-1 are the complementary angles to θ 1 and θ 2. These angles are labeled as

αand β.

Construct the vector
r 1 from point P to the focus F and by

using vector addition show with the aid of equation (7.3) that

r (t) +
r 1 =pˆe 1 or
r 1 = (p−t^2 / 4 p)ˆe 1 −tˆe 2

A unit vector in the direction of
r 1 is

ˆer 1 =(p−t

(^2) / 4 p)ˆe 1 −tˆe 2
√
(p−t^2 / 4 p)^2 +t^2
= (4p
(^2) −t (^2) )ˆe 1 − 4 ptˆe 2
√
(4p^2 −t^2 )^2 + 16 p^2 t^2

Using the definition of the dot product one can show

ˆet·ˆe 1 = cos α=√ t

4 p^2 +t^2

(−ˆet)·ˆer 1 = cos β=

√ −t(4 p^2 −t^2 ) + 8p^2 t

4 p^2 +t^2

√

(4 p^2 +t^2 )^2 + 16 p^2 t^2

=

√ t(t^2 + 4p^2 )

4 p^2 +t^2

√

(4 p^2 +t^2 )^2 + 16 p^2 t^2

If cos α= cos β for all values of the parameter t, then one must show that

√ t

4 p^2 +t^2

= t(t

(^2) + 4 p (^2) )
√
4 p^2 +t^2
√
(4p^2 +t^2 )^2 + 16p^2 t^2
(7 .4)

Using algebra one can establish that equation (7.4) is indeed true and so the angles

αand β are equal. Simplify the equation (7.4) to the form

√

(4 p^2 +t^2 )^2 + 16p^2 t^2 =t^2 + 4p^2

and then square both sides to show

16 p^4 − 8 p^2 t^2 +t^4 + 16 p^2 t^2 = (t^2 + 4p^2 )^2

which reduces to the identity (t^2 + 4p^2 )^2 = (t^2 + 4 p^2 )^2. The equality of the angles α

and β implies θ 1 =θ 2 or the angle of incidence is equal to the angle of reflection.

One can also show that the distances AF=FP which shows the triangle PFA is an

isosceles triangle with angle ∠F AP equal to angle ∠AP F implying the complementary

angles θ 1 and θ 2 are equal. These results show that all light coming in parallel to

the x−axis will be reflected by the mirrored parabolic surface and pass through the

focus. Conversely, if a light source is placed at the focus, than rays of light from the

focus are reflected parallel to the x−axis.