### Video Transcript

Suppose that we model an objectβs
trajectory in the solar system by a hyperbolic path in the coordinate plane, with
its origin at the sun and its units in astronomical units, Au. The π₯-axis is a line of symmetry
of this hyperbola. The object enters in the direction
of π¦ is three π₯ minus nine, leaves in the direction of π¦ is negative three π₯
plus nine, and passes within one Au of the sun at its closest point. Using the equations of the
asymptotes, find the equation of the objectβs path.

Weβre told that the objectβs path
is a hyperbola, that it enters in the direction of π¦ is three π₯ minus nine and
leaves in the direction of π¦ is negative three π₯ plus nine. So we can assume that these two
equations are the equations of the asymptotes of the hyperbola. In both of these equations, when π¦
is equal to zero, we find that π₯ is equal to three, so that the lines intersect at
the point with coordinates three, zero. Our first line has a slope of three
and a π¦-intercept at negative nine. And our second asymptote has a
slope of negative three and a π¦-intercept of plus nine.

We know that our object passes
within one Au of the sun at its closest point. This means that one of our vertices
must have coordinates one, zero, so that our hyperbola looks something like
this. And weβre told that the π₯-axis is
one of our lines of symmetry. If we take this is as our major or
transverse axis, then our hyperbola is horizontal or east westβopening. And weβre asked to use the
equations of the asymptotes to find the equation of the objectβs path. Now weβve established that the
center of the hyperbola is at the point three, zero. So this is a hyperbola in standard
position thatβs been shifted three units in the positive π₯-direction, where a unit
here is an astronomical unit.

Now we know that a hyperbola with
center β, π and with this orientation has the equation π₯ minus β squared over π
squared minus π¦ minus π squared over π squared is equal to one. And we concede that in our case, β
is equal to three and π is equal to zero. So that our equation will have the
form π₯ minus three squared over π squared minus π¦ squared over π squared is
equal to one. And all we need to do now is to
find the values of π and π. We can do this by using our
knowledge of hyperbolas.

Now we know that for any east
westβopening hyperbola, the slope of the asymptotes is positive and negative π over
π, where π and π are both positive. In our case, π over π is equal to
three, so that π is equal to three π. Just to make some room, Iβll move
things round a little bit. We have π equals three π and π₯
minus three squared over π squared minus π¦ squared over π squared is equal to
one. So now if we substitute π is equal
to three π into our equation, we have π₯ minus three squared over π squared minus
π¦ squared over nine π squared is equal to one.

Now we know that the hyperbola
passes through the point one, zero because this is our vertex. So we know that when π₯ is equal to
one, π¦ is equal to zero. And if we substitute these into our
equation, we have negative two squared over π squared is equal to one, so that π
squared is equal to four, and π is equal to two since π is positive. And remember that π is equal to
three times π, so π is equal to six. And the equation of the objectβs
path is π₯ minus three squared over four minus π¦ squared over 36 is equal to
one.