Calculus: Analytic Geometry and Calculus, with Vectors

6.4 Hyperbolas 387

then P traverses a branch of a hyperbola. Calculate the acceleration a and

show that a = k2r.

22 Copy Figure 6.48 and let a particle P move along the right branch of the

hyperbola in such a way that the vector r running from the origin to P is, at each

time t,

(1)

Show that the vector

(2)

r = (a cosh t)i + (b sinh t)j.

V = (a sinh t)i + (b cosh t)j

is a forward tangent to the hyperbola at P. Use the figure to show that, where

Xk=+1when k=land Xk=-1when k=2,

(3) FaP = (a cosh t + a2 + b2) i + (b sinh t)j

and then, with the aid of the simple fact that 1 + sinh2 t = cosh2 t, show that

(4) IFkPj = a2 + b2 cosh t + Xa.

Letting 4k be the angle which the vector FkP makes with the forward tangent
v at P, show that
FkP v
(5) IvI cos 'k =
I

F,kPI = /a2 + b2 sinh t

and hence that

(6) cos Ok =

V b

/

+

+ b2 y

4 + (a2 + b2)y2

Remark: The formula (6) shows that ¢1 = 02. This means that the lines drawn
from the foci to a point P on a hyperbola make equal angles with the tangent to
the hyperbola at P.
23 An ellipse and a hyperbola are confocal, that is, have the same foci.
Prove that they are orthogonal where they intersect.
Remark: One way to prove the result is to use the fact
that if F1 and F2 are the foci and P is a point of intersec-
tion of the ellipse and hyperbola, then the vectors F1P and
F2P make equal angles with the tangent to the hyperbola
at P and also make equal angles with the normal to the
ellipse at P. The miniature Figure 6.495 can help us
remember the fact.
24 The members of a family of confocal hyperbolas
have their foci at the points Fl(-1,0) and F2(1,O).

Figure 6.495

Sketch good approximations to graphs of six of them. Suggestion: Do not work
too long on an easy problem. Make a figure which tells you where the corners of
the handy boxes must be, and then sketch hyperbolas at the rate of two per
minute. Remark: There is a reason for knowing about these things. If the x axis
is an impenetrable barrier except for an opening on the interval -16 x < 1, your
curves are paths followed by particles of a liquid or gas (or perhaps by football
spectators) that stream through the gap.