### Video Transcript

In this video, weβll learn how to
write the equation of a hyperbola and find the hyperbolaβs vertices and foci using
its equation and graph. We call it the conic sections or
the curves we get when we intersect a cone with a plane. In the conic sections are the
ellipse, of which the circle is a special case, the parabola, and the hyperbola. And the hyperbola, which is the
curve weβre concerned with, consists of the two mirror-image curves resulting from a
plane intersecting both halves of the cone.

To give us an idea of what the
equation of the hyperbola represents, itβs helpful to compare the components of an
ellipse and a hyperbola. So letβs do that and then look more
closely at the equation of a hyperbola and work through some examples. Remember, an ellipse is a plane
curve defined by the equation π₯ squared over π squared plus π¦ squared over π
squared is equal to one. And for any point on the curve, the
sum of the distances from its two foci is a constant. And remember, a circle is a
particular type of ellipse where the two focal points are equal. A noncircular ellipse has two lines
of symmetry called the major and the minor axes. And the foci sit on the major axis
which is the longer of the two.

A hyperbola also has major and
minor axes, and these are, again, lines of symmetry. And again, the foci lie on the
major axis. But how an ellipse and a hyperbola
differ is that in defining the locus of points, thatβs the set of all points
satisfying its equation, itβs not the sum of the distances from the foci thatβs
constant, but the difference of the distances from the point to the two foci. The hyperbola we have here is in
whatβs called standard position. And thatβs where the center π is
at the origin, where π₯ is zero and π¦ is zero. Notice also that, in this case, the
π₯-squared term is positive. And weβd say in this case that the
hyperbola is horizontal or east-west opening; that is, the two branches are east and
west facing.

An east-west opening or a
horizontal hyperbola has its major axis parallel to the π₯-axis. And as weβve noted, that means the
π₯-squared term is positive. The hyperbola shown here is a north
southβopening hyperbola. And in this case, the π¦-squared
term is positive. So the branches face north and the
south, and itβs a vertical hyperbola. And in this case, the center is the
origin, so itβs in standard position. In a north southβopening hyperbola,
the major axis is parallel to the π¦-axis.

Now, weβre going to concentrate on
our east-west hyperbola, which has equation π₯ squared over π squared minus π¦
squared over π squared is equal to one, where π and π are greater than zero. The two points closest to the foci
on the major axis are called the vertices. And their coordinates are negative
π, zero and π, zero, so that the distance between the two vertices is two π. And the foci or foci are at the
points πΉ one is equal to negative π, zero and πΉ two is equal to π, zero. And π squared is equal to π
squared plus π squared.

Another important feature of the
hyperbola is the asymptotes. These are two lines which pass
through the center and which the branches get closer and closer to as π₯ gets larger
and larger, although the branches never touch the asymptotes. These are the lines π¦ is equal to
π over π π₯ and π¦ is equal to negative π over π times π₯. So, in fact, these are the lines
with slopes π over π and negative π over π. And remember, this is an east
westβopening hyperbola, which is centered at the origin. Itβs worth noting that for a north
southβopening hyperbola with equation π¦ squared over π squared minus π₯ squared
over π squared is equal to one, we need to be careful in redefining our vertices,
foci, and asymptotes.

In this case, our vertices of π
one is equal to zero, negative π and π two is zero, π. The foci have coordinates zero,
negative π and zero, π. And the asymptotes are the lines π¦
is equal to plus or minus π over π π₯. So now letβs look at how we might
derive the equation for the east westβopening hyperbola in standard position. And remember that for a hyperbola,
itβs the difference in distances from the foci to any point on the curve thatβs
constant. Now, π two is a point on our curve
with coordinates π, zero. So letβs take this as our point and
look at the distance of this from the two foci.

If we take π· one to be the
distance from the vertex π two to the first focal point πΉ one, thatβs equal to π
plus π. And now if we take π· two to be the
distance from π two to the second focal point πΉ two, thatβs equal to π minus
π. So now, if we take the difference
of these two distances, thatβs π plus π minus π minus π. Our πβs cancel each other out, so
weβre left with two π. And since the difference in
distances from the foci of any point is constant, this must be our constant. So now for any point π: π₯, π¦ on
the hyperbola, the difference between the distances from the two foci, π one minus
π two, is equal to positive or negative two π. And we need the positive or
negative because the sign of π one minus π two could be either, depending on which
branch the point sits.

From our graph, we see that we can
use the Pythagorean theorem to find π one and π two. If the base of our triangle with
hypotenuse π one is π plus π₯, then π one is the square root of π₯ plus π
squared plus π¦ squared. And if the base of our triangle
with hypotenuse π two is π minus π₯, then π two is the square root of π minus π₯
squared plus π¦ squared. And since π minus π₯ squared is
the same as π₯ minus π squared, and so π one minus π two is the square root of π₯
plus π squared plus π¦ squared minus the square root of π₯ minus π squared plus π¦
squared. And thatβs equal to positive or
negative two π.

Now, remember, weβre trying to
derive this equation, and this is gonna take a little bit of algebra. So Iβm going to make some space
here and swap sides for our expressions. Now, if we add this second term to
both sides, then on the left-hand side, this term cancels itself out. And now if we square both sides, we
have π₯ plus π squared plus π¦ squared on the left-hand side and our expanded
bracket on the right-hand side. And if we multiply out these two
brackets and collect like terms, we have π₯ times π minus π squared on the
left-hand side and positive or negative π times the square root of π₯ minus π
squared plus π¦ squared on the right-hand side.

So now just making some space and
rewriting that, if we square both sides again and expand our bracket inside the
square root, multiplying out the bracket on the right-hand side, we can eliminate
negative two π squared π₯π on both sides by adding two π squared π₯π. And by taking π raised to the
fourth power to the right-hand side and π squared π₯ squared and π squared π¦
squared to the left-hand side, this reduces to π squared minus π squared times π₯
squared minus π squared π¦ squared is equal to π squared times π squared minus π
squared. And now dividing everything by π
squared times π squared minus π squared, we can make quite a few
cancellations. And weβre left with π₯ squared over
π squared minus π¦ squared over π squared minus π squared is equal to one.

And now, if we set π squared equal
to π squared minus π squared, weβre left with π₯ squared over π squared minus π¦
squared over π squared is equal to one. And this is the equation of a
horizontal or east westβopening hyperbola in standard position. Now, remember, the hyperbola has
two asymptotes, π¦ is plus or minus π over π times π₯, which it approaches as π₯
gets large but never quite touches. We can see how this works with
respect to our equation by solving for π¦ and looking at the behavior as π₯
approaches positive and negative β. And in fact, making π¦ the subject
of the equation, we have π¦ is equal to the positive or negative square root of π
squared over π squared π₯ squared minus π squared.

We note first that the argument of
the square root is positive since π₯ squared is always greater than or equal to π
squared. And as π₯ approaches positive or
negative β, the π squared term becomes insignificant. Hence, π¦ approaches the positive
or negative square root of π squared over π squared times π₯ squared, which is
positive or negative π over π times π₯. That is the straight lines through
the origin with slopes positive or negative π over π. So now letβs look at an example of
how to graph a hyperbola in standard position.

Given the hyperbola defined by the
equation π₯ squared over 36 minus π¦ squared over 16 is equal to one, find the
vertices, the foci, and the asymptotes, and use these to graph the hyperbola.

Weβre given the equation of a
hyperbola, π₯ squared over 36 minus π¦ squared over 16 is equal to one, which is in
the form π₯ squared over π squared minus π¦ squared over π squared is equal to
one, and where π and π are both positive. In our case then, π squared is 36,
and π squared is 16. And so π is six, and π is
four. For a hyperbola of this form, the
center is at the origin zero, zero. If we look again at our equation,
we can see that the positive term is π₯ squared and the negative term is π¦
squared. And this means that the major or
transversed axis is on the π₯-axis. And remember that both the foci and
the vertices are on the major axis, so that our hyperbola will look something like
this.

Remember that the vertices are the
points on each branch which are closest to the center. And for a hyperbola of this type,
the vertices are at negative π, zero and π, zero, so that in our case, the
vertices have coordinates negative six, zero and six, zero. The coordinates of the foci or foci
πΉ one and πΉ two are negative π, zero and π, zero, where π squared is π squared
plus π squared. So in our case, π squared is 36
plus 16, which is 52. And therefore, π which is positive
is the square root of 52. And to one decimal place, thatβs
7.2, so that our foci have coordinates negative 7.2, zero and 7.2, zero.

The asymptotes of a hyperbola in
standard position are the lines π¦ is π over π times π₯ and π¦ is negative π over
π π₯. So if we call our asymptotes π΄ one
and π΄ two with π equal to six and π equal to four, we have π΄ one equal to four
over six π₯, which is two over three π₯. And our second asymptote is
negative two over three times π₯. And now we have all the information
we need to graph our hyperbola. We have the vertices with
coordinates negative six, zero and six, zero; the foci with coordinates negative
7.2, zero and 7.2, zero; and our two asymptotes. Thatβs π¦ is two over three π₯ and
negative two over three π₯.

Now letβs look at an example of how
we find the equation of a hyperbola from its asymptotes in a real-world context.

A gardener wants to grow a hedge in
the shape of a hyperbola with a fountain at its center. At its closest, the hedge will be a
distance of 20 yards from the fountain. The fountain is at the origin of
the coordinate system with units in yards, and the hyperbola traced by the hedge has
asymptotes π¦ is three over four π₯ and π¦ is negative three over four π₯. Find the equation of the
hyperbola.

To answer this question, letβs
begin by sketching our hyperbola using the information given. Weβre told that the fountain is at
the center, but also that the fountain is the origin of the coordinate system. The hedge will be the hyperbolaβs
branches, which will approach the asymptotes π¦ is three over four π₯ and negative
three over four π₯ as π₯ gets larger in the positive and negative directions.

Our question says, at its closest,
the hedge will be a distance of 20 yards from the fountain. And since the fountain is the
origin of our coordinate system, we can take the points with coordinates 20, zero
and negative 20, zero as our vertices. This means that weβve opted for an
east westβopening hyperbola. And with the center at the origin,
this has an equation of the form π₯ squared over π squared minus π¦ squared over π
squared is equal to one.

And we know that for an east
westβopening hyperbola, the asymptotes are π¦ is positive or negative π over π
times π₯. Weβre given that these are three
over four and negative three over four π₯. And so itβs possible that π is
equal to three and π is equal to four. Our asymptotes are the lines π¦ is
equal to positive or negative three over four π₯, which are the lines with the
slopes positive and negative three over four. However, our fractions couldβve
been simplified, and π and π could actually be larger than these numbers. However, we do know somewhere else
where π appears. And thatβs on the transverse or
major axis of the hyperbola.

We know that the distance from the
center of the hyperbola to the vertex is π since the vertices of such a hyperbola
have coordinates plus or minus π, zero. So in our case, π is equal to
20. And we know that our positive slope
is π over π, and thatβs three over four. And since π is 20, that means
three over four is equal to π over 20. If we cross multiply, we have 20
times three over four is equal to π. And since four divides 20 five
times, our π is equal to 15.

And now, since we have the value of
π and the value of π, we can substitute these into our hyperbola equation where π
squared is 225 and π squared is 400. And our hyperbola is π₯ squared
over 400 minus π¦ squared over 225 is equal to one, so that our hedge is in the
shape of a hyperbola with equation π₯ squared over 400 minus π¦ squared over 225 is
equal to one.

Now, so far, weβve been looking at
hyperbolas which were centered at the origin. But of course, they could be
centered anywhere in the π₯π¦-plane. If we translate our hyperbola in
the plane so that its center has coordinates β, π, our translated hyperbola has the
equation π₯ minus β squared over π squared minus π¦ minus π squared over π
squared is equal to one. In this form, since the π₯-term is
positive, the major axis is parallel to the π₯-axis. And remember, the major axis
contains the vertices and the foci. And this is an east westβopening
hyperbola or a horizontal hyperbola.

Itβs worth noting that if the
π¦-term was positive, then our hyperbola would look like this with the transverse or
major axis parallel to the π¦-axis. And this is called a north
southβopening or a vertical hyperbola. So now letβs see how this
translation of the center of the hyperbola works in practice.

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.

Letβs conclude our look at the
equation of a hyperbola by recapping some key points. A hyperbola in standard position
has its center at zero, zero. The orientation of the hyperbola is
determined by whether itβs the π₯ squared or the π¦ squared term that has the
positive sign. A hyperbola with center β, π has
been shifted or translated in the plane β units in the π₯-direction and π units in
the π¦-direction.

The vertices are the two points
which are the closest to the center on the branches. The foci are the points a distance
π from the center such that for any point on a branch of the hyperbola, the
difference in the distances from the point to the two foci is the constant two
π. The asymptotes are the intersecting
lines through the center of the hyperbola, which the branches of the hyperbola
approach as π₯ gets large but never touch.