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Parametric Equations and Polar Coordinates

In this section , we extend the concepts of calculus to curves described by parametric equations and polar coordinates. For instance, in order to study the motion of an object such as an airplane in two dimensions, we would need to describe the objects position (x,y) as a function of the parameter t(time). That is , we write the position in the form , where and are functions to which our existing techniques of calculus can be applied. In addition, we will explore how to use polar coordinates to represent curves, not as a set of points , but rather , by specifying the points by the distance from the origin to the point. Polar coordinates are especially convenient for describing circles, such as those that occur in propagating sound waves.

Plane Curves and Parametric equations

It is often convenient to describe the location of appoint in the plane in terms of a parameter. For instance, in tracking the movement of a satellite , we would naturally want to give its location in terms of time. In this way , we not only know the path it follows, but we also know when it passes through each point.

Given any pair of functions and defined on the same domain D, the equations

, are called parametric equations. Notice that for each choice of t, the parametric equations specify a point in the xy-plane. The collection of all such points is called the graph of the parametric equations. In the case where and are continuous function and D is an interval of the real line, the graph is a curve in the xy-plane, referred to as a plane curve.

The choice of the letter (t) to denote the independent variable ( called the parameter) should make you think of time, which is often what the parameter represents.

Example

Find parametric equations for the line segment joining the points and .

Solution

For a line segment, the parametric equation can be linear functions;

,

Where a,b,c,d are some constants to be determined

at at

Hence where

is a pair of parametric equations determining the line segment

Example

Find parametric equation of the curve :

Solution

Any equation of the form can be converted to parametric form by letting the parameter . then we have the equations:

using these parametric equations to graph the curve, we obtain

It would be possible to solve the given equation

( ) for x as four functions of y and graph them individually, but the parametric equations provide a much easier method.

In general , if we need to graph an equation of the form , we can use the parametric equations:

Example

Sketch the plane curve defined by the parametric equations

Solution

In the following table we list a number of values of the parameter t and corresponding values of x and y.

txy

-22-1

-15-1/2

060

151/2

221

3-33/2

4-102

We have plotted these points and connected them with a smooth curve as in figure below

It is also possible to eliminate the parameter here, by solving for t in terms of y. We have , so that

. The graph of this last equation is a parabola opening to the left. However, the plane curve we are looking for is the portion of this parabola corresponding to . From the table notice that this corresponds to , so that the plane curve is the portion of the parabola indicated in the figure above, where we have also indicated a number of points on the curve.

Example

Find the path of a projectile thrown horizontally with initial speed of 20 ft/s from a height of 64 feet.

Solution

The path is defined by the parametric equations

Where t represents time (in seconds). This describes the plane curve shown in the figure below

Note that the orientation indicated in the graph gives the direction of motion. Although we could eliminate the parameter here , the parametric equations provide us with more information, as they tell us when the object is located at a given point and indicate the direction of motion. We indicate the location of the projectile at several times in figure shown above.

Example

Sketch the plane defined by the parametric equations:

Solution

(a)

the default graph produced by most graphing calculators looks something like the curve shown in the following figure

We have added arrows indicating the orientation. The curve is looks like an ellipse, but with some more thought we can identify that it is a circle. Rather than eliminate the parameter by solving for t in terms of either x or y, instead notice that :

So the plane curve lies on the circle of radius 2 centered at the origin. In fact, it is the whole circle, as we can see by recognizing what the parameter represents in this case. Recall that from the definition of sine and cosine that if is a point on the unit circle and is the angle from the positive x-axis to the line segment joining and the origin, then we define and . Since we have and , the parameter t corresponds to the angle . The curve is the entire circle of radius 2, traced out as the angle t ranges from 0 to 2

(b)

What would change if the domain were limited to

? Since we have identified t as the angle as measured from the positive x-axis, it should be clear that we will now get the top half of the circle, as shown in the figure below:

Example

Identify the plane curves

(a)

(b) (c) all for

Solution

Using a computer-generated sketch of (a) gives us the following figure:

It is difficult to determine the sketch whether the curve is an ellipse or simply a distorted graph of a circle. You can rule out a circle, since the parametric equations produce x-values between -2 and 2 and y-values between -3 and 3. To verify that this is an ellipse , observe that

A computer sketch of (b) is shown in the following figure

We can verify that this is a circle :

Finally the sketch of ( c) is shown below

We can verify that this is a circle

Example

Find parametric equations for the portion of the parabola from

Solution

Any equation of the form can be converted to parametric form simply by defining . Here, this gives us , so that

is a parametric representation of the curve.

Example

Sketch the plane curves:

(a) (b)

Solution

Since there is no restriction places on t, we can assume that t can be any real number. Eliminating the parameter in (a), we get , so that the parametric equations in (a) correspond to the parabola

shown in the figure below

Notice that the graph includes the entire parabola, since t and hence, can be any real n umber.

For (b) when we eliminate the parameter, we get and so, , this gives the same parabola as in (a) . However, the initial sketch of the parametric equations shown in the following figure shown only the right half of the parabola

To verify that this is correct, note that since , we have that for every real number t. Therefore , the curve is only the right half of the parabola

as shown.

Example

Sketch the plane curves:

(a)

(b) Solution

(a) The sketch is shown in the figure below

From the vertical line test, this is not the graph of any function. Further, converting to an x-y equation here is messy and not particularly helpful. However, examine that parametric equations to see if important portions of the graph have been left out ( e.g., there supposed to be anything to the left of ?). Here, for all and has no maximum or minimum . It seems that most of the graph is indeed shown in the last figure.

(b) the sketch is shown below

Once again , this is not a familiar graph. To get an idea of the scope of the graph, note that has no maximum or minimum . To find the minimum of

, note that critical numbers are at and with corresponding function values 4 and -9/4 , respectively. We should conclude that , as shown in the figure for this point.

Example

Suppose that a missile is fired toward your location from 500 miles away and follows a flight path given by the parametric equations:

Two minutes later , you fire an interceptor missile from your location following the flight path :

Determine whether the interceptor missile hits its target.

Solution

In the following figure :

We have plotted the flight paths for both missiles simultaneously. The two paths clearly intersect, but this does not necessarily mean that the two missiles collide. For that to happen, they need to be at the same point at the same time. To determine whether there are any values of t for which both paths are simultaneously passing through the same point, we set the two x-values equal:

And obtain one solution : . Notice that this simply says that the two missiles have the same x-coordinate when . Unfortunately, the y-coordinates are not the same here, since when , we have

You can see this graphically by plotting the two paths simultaneously for only, as we have done in the figure below:

From the graph, you can see that the two missiles pass one another without colliding. So, by the time the interceptor missile intersects the flight path of the incoming missile, it is long gone , so the interceptor missile does not hit its target.

Calculus and Parametric equations

Tangents

Our initial aim is to find a way to determine the slopes of tangent lines to the curves that are defined parametrically. First, recall that for a differentiable function , the slope of the tangent line at the point is given by .Written in Leibniz notation, the slope is . Both x and y are functions of the parameter t. Notice that if and both have derivatives that are continuous at t=c, the chain rule gives us:

as long as , we have then

Where . In the case where , we define

provided the limit exists.

We can use (2.1) to calculate second ( as well as higher order) derivatives. Notice that if we replace y by , we get

ExampleA curve C is defined by the parametric equations .(a) Show that C has two tangents at the point and find their equations.

(b)Find the point on C where the tangent is horizontal or vertical.

(c)Determine where the curve is concave upward or downward.

(d) Sketch the curve.

Solution

(a) Notice that when or . Therefore, the point on C arises from two values of the parameter, . This indicated that C crosses itself at (3,0). Since

The slope of the tangent when is , so the equations of the tangents at (3,0) are

(b) C has a horizontal tangent when , that is when and . Since , this happens when , that is . The corresponding points on C are and . C has a vertical tangent when , that is , ( Note that ). The corresponding point on C is .

(c) To determine concavity we calculate the second derivative:

Thus , the curve is concave upward when () and concave downward when ().(d) The figure is as following: Example

Find the slope of the tangent to the path , at (a) ; (b) , and

(c ) the point .

Solution

(a) First , note that;

From (2.1) , the slope of the tangent line at is then

(b) The slope of the tangent line at is:

EMBED Equation.3 ( c) To determine the slope at the point , we must first determine a value of t that corresponds to the point in this case, notice that gives and . Here, we have

and consequently, we must use (2.2) to compute . Since the limit has indeterminate form () , we use 1Hopitals Rule, to get

Which does not exist, since the limit in the numerator is 6 and the limit in the denominator is 0. This says that the slope of the tangent line at is undefined.

The tangent lines are shown in the figure below for different t-values. Notice that the tangent line at the point is vertical

Finding slopes of tangent lines can help us identify many points of interest.

Example

Identify all points at which the plane curve has a horizontal or vertical tangent line.

Solution

A sketch of the curve is shown in the figure below:

There appear to be two locations ( the top and bottom of the bow) with horizontal tangent lines and one point ( the far right edge of the bow) with a vertical tangent line. Recall that horizontal tangent lines occur where .

From (2.1), we then have , which can occur only when provided that for the same value of t. Since only when is an odd multiple of , we have that , only when and so , . The corresponding points on the curve are then:

and

Notice that and reproduce the first and the third points, respectively, and so on. The points and are on the top and bottom of the bow , respectively, where there clearly are horizontal tangents. The points and should not seem quite right, though. These points are on the extreme ends of the bow and certainly dont look like they have vertical or horizontal tangents.

To find points where there is a vertical tangent, we need to see where but . Setting , we get , which occurs if or The corresponding points are

Since , for or , there is a vertical tangent line only at the point .

Theorem (2.1)

Suppose that are continuous. Then for the curve defined by the parametric equations and ,

(i) if and , there is a horizontal tangent line at the point ,

(ii) if and , there is a vertical tangent line at the point .Recall that if the position of an object moving along straight line is given by the differentiable function f(t), the objects velocity is given by . The situation with parametric equations is completely similar. If the position is given by , for differentiable functions and , then the horizontal component of velocity is given by and the vertical component of velocity is given by ( see the figure below)

The speed can be defined to be . From this, note that the speed is 0 if and only if . In this event, there is no horizontal no vertical motion.

Example

For the path of scrambler , , find the horizontal and vertical component of velocity and speed at times and , and indicate the direction of motion. Also determine all times at which the speed is zero.

Solution

Here, the horizontal component of velocity is

and the vertical component is

at t=0, the horizontal and vertical components of velocity both equal 2 and the speed is . The rider is located at the point and is moving to the right [ since ]and ( vertical) and the speed is . At this time, the rider is located at the point and is moving to the left [ since ].

In general, the speed of the rider at time t is given by

Using the identities , and , the speed is 0 whenever

. This occurs when ,

or .

The corresponding points on the curve are , , and . These points are the three tips of the path seen in the figure below;

The Area

Referring to the last figure notice that the path begins and ends at the same point and so, encloses an area. An interesting question is to determine the area enclosed by such curve. Computing areas in parametric equations is a straightforward extension of the original development of integration. Recall that for a continuous function f defined on [a,b], where on [a,b], the area under the curve for is given by :

Now, suppose that this same curve is described parametrically by and , where the curve is traversed exactly once for . We can then compute the area by making the substitution . It then follows that and so, the area is given by

,

Where it is clear that we have changed the limits of integration to match the new variable of integration. We generalize this result in the following theorem

Theorem (2.2)

Suppose that the parametric equations and , with , describe a curve that is traced out clockwise exactly once, as t increases from c to d and where the curve does not intersect itself, except that the initial and terminal points are the same [ i.e., and ]. Then the enclosed area is given by

If the curve is traced out counterclockwise, then the enclosed area is given by

The new area formulas given in theorem (2.2) turn to be quite useful. We can use these to find the area enclosed by a parametric curve.

Example :

Determine the area under the parametric curve given by ; ,

Solution;

the area is then:

The parametric curve used in the previous example is called a cycloid. Its general form is :

it is like a wheel of radius r that is rolling to the right and we are tracing the path of a point p that is initially touching the ground. The point p will be back on the ground after it has rotated ( ).

Example

Find the area enclosed by the path of the scrambler Solution

Notice that the curve is traced out counterclockwise once for . From (2.5), the area is then

Where the integration has been evaluated using a CAS.

Example

Find the area by the ellipse

( for constants ).

Solution

One way to compute the area is to solve the equation for y to obtain

and then integrate:

This integration also can be evaluated by trigonometric substitution or by using a CAS, but a simpler, more elegant way to compute the area is to use parametric equations . Notice that the ellipse is described parametrically by , for . The ellipse is then traced out counterclockwise exactly once for , so that the area is given by (2.5) to be

Where this last integral can be evaluated by using the half-angle formula:

Arc Length and Surface Area

In this section we investigate arc length and surface area for curves defined parametrically. Along the way, we explore one of the most famous and interesting curves in mathematics.

Let C be the curve defined by the parametric equations and , for ( see the figure below)

Where ,, and are continuous on the interval . We further assume that the curve does not intersect itself, except possibly at a finite number of points. Our goal is to compute the length of the curve ( the arc length). We begin by constructing an approximation.

First , we divide the t-interval into n subintervals of equal length, :

Where , for each

For each subinterval , we approximate the arc length of the portion of the curve joining the point to the point with the length of the line segment joining these points. This approximation is shown in the figure below:

for the case n=4. We have

Recall that from the mean value theorem , we have that :

, and

Where and are some points in the interval . This gives us :

Notice that is small, then and are close together. So, we can make the further approximation

Taking the limit as then gives us the exact arc length which we should recognize as an integral:

We summarize this discussion in the following theorem.

Theorem (3.1)

For the curve defined parametrically by ,

, if are continuous on and the curve does not intersect itself ( except possibly at a finite number of points), then the arc length s of the curve is given by

Example

Find the arc length of the scrambler curve , for . Also, find the average speed of the scrambler over this interval.

Solution

The curve is shown in the following figure:

First, note that are all continuous on the interval [ 0, 2]. From (3.1), we then have

Since , where we have approximated the last integral numerically. To find the average speed over the given interval, we simply divide the arc length ( i.e., the distance traveled), by the total time, , to obtain

Theorem (3.1) allows the curve to intersect itself at a finite number of points, but a curve cannot intersect itself over an entire interval of values of the parameter t. To see why this requirement is needed, notice that the parametric equations , for , describe the circle of radius 1 centered at the origin. However, the circle is traversed twice as t ranges from 0 to 4 . If we apply theorem (3.1) to this curve , we would obtain:

Which corresponds to twice the arc length ( circumference) of the circle. As we can see, if a curve intersects itself over an entire interval of values of t, the arc length of such a portion of the curve is counted twice by (3.1).ExampleDetermine the length of the parametric curve given by :

Solution

This is a circle of radius 3 centred at the origin . also it is traversed only once in this range .

Hence , the length is :

since this is a circle, length is equal to circumference

( )

on the other hand,. If we use:

then: , and the length is:

Example

An 8-foot-tall ladder stands vertically against a wall. The bottom of the ladder is pulled along the floor , with the top remaining in contact with the wall , until the ladder rests flat on the floor. Find the distance traveled by the midpoint of the ladder.

Solution

We first find parametric equations for the position of the midpoint of the ladder. We orient the x- and y-axes as shown in the figure below:

Let x denote the distance from the wall to the bottom of the ladder and let y denote the distance from the floor to the top of the ladder. Since the ladder is 8 feet long, observe that . defining the parameter , we have . The midpoint of the ladder has coordinates ( )and so , parametric equations for the midpoint are

When the ladder stands vertically against the wall, we have and when it lies flat on the floor, . So . The arc length is then given by

Substituting gives us or . For the limits of integration, note that when and when . The arc length is then

Surface Area

We can use the arc length formula to find a formula for the surface area of a surface of revolution. Recall that if the curve for is revolved about the x-axis ( see the figure below)

The surface area is given by :

Let C be the curve defined by the parametric equations and with , where are continuous and where the curve does not intersect itself for . The corresponding formula for surface area with parametric equations:

More generally, we have that if the curve is revolved about the line , the surface area is given by

Likewise, if we revolve the curve about the line x=d, the surface area is given by:

So the surface area has some thing common for all formulas , that is

Example

Find the surface area formed by revolving the half-ellipse about the x-axis ( see the figure)

Solution

It would be a mess to set up the integral for Instead , notice that we can represent the curve by the parametric equations , for . From the formula for surface area we have :

Where we used a CAS to evaluate the integral.

Example

Determine the surface area of the solid obtained by rotating the following parametric curve about the x-axis

Solution

First, evaluate the derivatives:

then, evaluate as following:

we can drop the sign of absolute value since both sine and cosine are positive in the given range of

The surface area is then ;

Let , then

Example

Find the surface area of the surface formed by revolving the curve , for , about the line .

Solution

A sketch of the curve is shown in the figure below:

Since x-values on the curve are all less than 2 , the radius of the solid of revolution is and so, the surface area is given by

where we have approximated the value of the integral numerically.

Polar Coordinates

A coordinate system represents a point in the plane by an ordered pair of numbers called coordinates. Usually we use Cartesian coordinates, which are directed distances from two perpendicular axes. Here we describe a coordinate system introduced by Newton, called the polar coordinate system, which is more convenient for many purposes.

We choose a point in the plane that is called the pole ( or origin) and is labeled 0. Then we draw a ray ( half-line) straight at 0 called the polar axis. This axis is usually drawn horizontally to the right and corresponds to the positive x-axis in Cartesian coordinates.

If p is any other point in the plane , let r be the distance from 0 to p and let be the angle ( usually measured in radius) between the polar axis and the line OP as in the figure below

Then the point P is represented by the ordered pair ( ) and r , are called polar coordinates of P. we use the convention that an angle is positive if measured in the counterclockwise direction from the polar axis and the negative in the clockwise direction. If , then and we agree that represents the pole for any value of .

We extend the meaning of polar coordinates () to the case in which r is negative by agreeing that, as in the figure below , the points () and () lie on the same line through 0 and at the same distance from 0, but on opposite sides of 0.

If , the point () lies in the same quadrant as ; if , it lies in the in the quadrant on the opposite side of the pole. Notice that () represents the same point as

().

Example

Plot the points whose polar coordinates are given :

(a) (b)

( c)

(d)

Solution

The points are plotted in the following figures:In part (d) the point is located three units from the pole in the fourth quadrant because the angle is in the second quadrant and is negative.

In the Cartesian coordinate system every point has only one representation, but in the polar coordinate system each point has many representations. For instance, the point in the above example could be written as

or or

(see the figures below)

In fact, since a complete counterclockwise rotation is given by an angle 2, the point represented by polar coordinates is also represented by :

where n is an integer.

The connection between polar and Cartesian coordinates can be seen in the following figure

In which the pole corresponds to the origin and the polar axis coincides with the positive x-axis. If the point P has Cartesian coordinates and the polar coordinates , then, from the figure , we have

and also

Although equations (1) were deduced from the last figure, which illustrates the case where and , these equations are valid for all values of r and .

Equations (1) allows us to find the Cartesian coordinates of a point when the polar coordinates are known. To find and when x and y are known, we use the equations

Which can be deduced from equations (1) or simply read from the last figure.

Example

Convert the point from polar to Cartesian coordinates.

Solution

Since and , equation (1) give

Therefore, the point is in Cartesian coordinates.

Example

Represent the point with Cartesian coordinates in terms of polar coordinates.

Solution

If we choose r to be positive, then equation (2) give

Since the point lies in the fourth quadrant, we can choose or . Thus , one possible answer is ; another is .

Polar Curves

The graph of a polar equation , or more generally , consists of all points that have at least one polar representation whose coordinates satisfy the equation.

Example

What curve is represented by the polar equation ?

Solution

The curve consists of all points with . Since represents the distance from the point to the pole, the curve represents the circle with center 0 and radius 2. In general, the equation represents a circle with center 0 and radius . ( see the figure)

Example

Sketch the curve .

Solution

This curve consists of all points such that the polar angle is 1 radian. It is the straight line that passes through 0 and makes an angle of 1 radian with the polar axis ( see the figure)

Notice that the points on the line with are in the first quadrant, whereas those with are in the third quadrant.

Example

(a) Sketch the curve with polar equation Find a Cartesian equation for this curve.

Solution

(a) In the following figure we find the values of r for some convenient values of and plot the corresponding points .

Then we join these points to sketch the curve, which appears to be a circle. We have used only values of between 0 and , since if we let increase beyond , we obtain the same points again.

(b) To convert the given equation into Cartesian equation we use equation (1) and (2) . From we have , so the equation becomes , which gives

or

Completing the square, we obtain

which is an equation of a circle with center and radius 1. The following figure shows the geometrical illustration that the circle has the equation . The angle is a right angle and so

Example

Sketch the curve .Solution

Instead of plotting the points as in the previous example , we first sketch the graph of in Cartesian coordinates as in the figure below by shifting the sine curve up one unit.

This enables us to read at a glance the values of r that correspond to increasing values of . For instance , we see that as increases from 0 to , ( the distance from 0) increases from 1 to 2, so we sketch the corresponding part of the polar curve in the following figures

In (a) as increases from 0 to , increases from 1 to 2, so we sketch the next part to the curve as in (b). We see that as increases from to , decreases from 2 to 1. As increases from to 3 , decreases from 1 to 0 as in part ( c). Finally, as increases from to , r increases from 0 to 1 as in part (d).If we let increase beyond or decrease beyond 0, we would simply retrace our path. Putting together the parts of the curve from figures (a) to (d) , we sketch the complete curve in part (e). It is called a cardioid because its shaped like a heart.

Example

Sketch the curve .

Solution

We first sketch , , in Cartesian coordinates in the following figure:

As increases from to , r goes from 0 to -1 as in the part (1) in the following figure

This means that the distance from 0 increases from 0 to 1, but instead of being in the first quadrant this portion of the polar curve ( indicated by 2) lies on the opposite side of the pole in the third quadrant. The remainder of the curve is drawn in a similar fashion, with the arrows and numbers indicating the order in which the portion are traced out. The resulting curve has four loops and is called a four-leaved rose.

Areas and lengths in polar coordinates

In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the area of a sector of a circle

where, as in the figure below , r is the radius and is the radian measure of the central angle.

The formula above follows from the fact that the area of a sector is proportional to its central angle:

Let R be the origin, illustrated in the figure below, bounded by the polar curve and by the rays and , where f is a positive continuous function and where .

We divide the interval into subintervals with endpoints and equal width . The rays then divide into smaller regions with central angle . If we choose in the i-th subinterval , then the area of the ith region is approximated by the area of the sector of a circle with central angle and radius

EMBED Equation.3 ( see the figure below)

Thus from formula (1) we have

and so an approximation to the total area of is

It appears from the last figure that the approximation in formula (2) improves as . But the sums in formula (2) are Riemann sums for the function ,so

It therefore appears plausible and can be proved that the formula for area A of the polar region R is

Formula (3) is often written as

with the understanding that .Example

Find the area enclosed by one loop of the four-leaved rose .

Solution

The curve was sketched in one of previous examples as in the figure below

Notice that the region enclosed by the right loop is swept out by a ray that rotates from to . Therefore , formula (4) gives

Example

Find the area of the region that lies inside the circle

and outside the cardioid .

Solution

The cardioid and the circle are sketched in the following figure and the desired region is shaded.

The values of a and b in formula (4) are determined by finding the points of intersection of the two curves. They intersect when , which gives , so . The desired area can be found by subtracting the area inside the cardioid between and from the area inside the circle from to . Thus

Since the region is symmetric about the vertical axis , we can write

The last example illustrates the procedure for finding the area of the region bounded by two polar curves . in general , Let be a region, that is bounded by curves with polar equations , , and , where and ,as illustrated in the following figure

The area A of R is found by subtracting the area inside from the area inside , so using formula (3) we have

Conic Sections

In this section we give geometric definitions of parabolas, ellipses, and hyperbolas and derive their standard equations. They are called conic sections, or conics, because they result from intersecting a cone with a plane as shown in the following figures:

Parabolas

A parabola is the set of points in a plane that are equidistant from a fixed point F ( called the focus) and a fixed line ( called the directrix) . This definition is illustrated by the following figure:

Notice that the point halfway between the focus and directrix lies on the parabola; it is called vertex . The line through the focus perpendicular to the directrix is called the axis of the parabola.

We obtain a particularly simple equation for a parabola if we place its vertex at the origin 0 and its directrix parallel to the x-axis as in the figure below:

If the focus is the point , then the directrix has the equation . If is any point on the parabola, then the distance from p to the focus is:

and the distance from p to the directrix is :

the last figure illustrates the case where . The defining property of a parabola is that these distances are equal:

we get an equivalent equation by squaring and simplifying:

An equation of parabola with focus ( 0,p) and directrix y=-p is x2 = 4py (1)

If we write , then the standard equation of a parabola (1) becomes . It opens upward if

and downward if . The figures below showing different cases:

If we interchange x and y in (1) we obtain :

Which is an equation of the parabola with focus and directrix .

The parabola opens to the right if and to the left if ( see the figure below)

In these both cases the graph is symmetric with respect to x-axis, which is the axis of the parabola.

Example

Find the focus and the directrix of the parabola and sketch the graph.

Solution

If we write the equation and compare it with equation (2) , we see that , so . Thus, the focus is and the directrix is .

The sketch is shown in the following figure:

Ellipses

An ellipse is the set of points in a plane the sum of whose distances from two fixed points F1 and F2 is a constant ( see the figure )

These two fixed points are called the foci ( plural of focus) .

In order to obtain the simplest equation for an ellipse , we place the foci on the x-axis at the points and as in the figure;

So that the origin is halfway between the foci. Let the sum of the distances from a point on the ellipse to the foci be .Then is a point on the ellipse when :

squaring both sides , we have

which simplifies to :

we square again:

which becomes :

From triangle in the last figure we see that

, so and , therefore, .

For convenience , let . then the equation of the ellipse becomes . If both sides are divided by we obtain:

Since , it follows that . The x-intercepts are found by setting . Then , or , so .

The corresponding points and are called the vertices of the ellipse and the line segment joining the vertices is called the major axis . To find the y-intercepts we set and obtain , so . equation (3) is unchanged if is replaced by or is replaced by , so the ellipse is symmetric about both axes . Notice that if the foci coincide, then , so and the ellipse becomes a circle with radius .

We summarize the discussion as follows

The ellipse has foci

, where , and vertices ( 4 )

see the figure below

If the foci of an ellipse are located on the y-axis at

, then we can find its equation by interchanging x and y in (4)

The ellipse has foci

where and vertices (5)

see the figure below

Example

Sketch the graph of and locate the foci.

Solution

Divide both sides of the equation by 144 :

The equation is now in the standard form for an ellipse , so we have , , , . The x-intercepts are , and the y-intercepts are . Also ,

, so and the foci are . the graph is sketched in the following figure:

Example

Find an equation of the ellipse with foci and vertices .Solution

Using the notation of (5) , we have and . Then we obtain , so the equation of the ellipse is :

Another way of writing the equation is

HyperbolasA hyperbola is the set of all points in a plane the difference of whose distances from two fixed points F1 and F2 ( the foci ) is a constant. This definition is illustrated in the following figure :

Notice that the definition of the hyperbola is similar to that of an ellipse : the only change is that the sum of distances has become a difference of distances. In fact , the derivation of the equation of a hyperbola is also similar to the one given earlier for an ellipse. hence the foci are on the x-axis at and the difference of the distance is : , then the equation of the hyperbola is :

EMBED Equation.DSMT4 where . Notice that the x-intercepts are again , and the points and are the vertices of the hyperbola. But if we put in equation (6) we get , which is impossible , so there is no y-intercept. The hyperbola is symmetric with respect to both axes.

To analyse the hyperbola further , we look at equation 6 and obtain

this shows that .

Therefore , we have . This means that the hyperbola consists of two parts, called the branches.

When we draw a hyperbola it is useful to first draw its asymptotes, which are the dashed lines :

shown in the figure below:

Both branches of the hyperbola approach the asymptotes; that is , they come arbitrarily close to the asymptotes.

The hyperbola has foci , where , vertices , and asymptotes (7)

If the foci of a hyperbola are on the y-axis, then by reversing the roles of x and y we obtain the following information , which is illustrated in the following figure:

The hyperbola has foci , where , vertices and asymptotes (8)

Example

Find the foci and asymptotes of the hyperbola and sketch its graph.

Solution

If we divide both sides of the equation by 144, it becomes : which is of the form given in (7) with and . Since , the foci are . The asymptotes are the lines . The graph is shown in the figure below:

Example

Find the foci and equation of hyperbola with vertices and asymptotes . Solution

From (8) and the given information , we see that a=1 and . Thus . The foci are and the equation of the hyperbola is

Shifted Conics

We can shift conics by taking the standard equations discussed in (1),(2),(4),(5),(7),and (8) and replacing x and y by ( x-h) and (y-k).

Example

Find an equation of the ellipse with foci , and vertices ,.

Solution

The major axis is the line segment that joins the vertices , and has the length 4 , so . The distance between the foci is 2, so . Thus , . Since the center of the ellipse is , we replace and in (4) by and to obtain:

as the equation of the ellipseExample

Sketch the conic and find its foci

Solution

We complete the squares as follows:

This is in form (8) except that and are replaced by and . Thus ,

The hyperbola is shifted four units to the right and one unit upward. The foci are and the vertices are and . The asymptotes are . The hyperbola is sketched in the following figure :

EMBED Equation.3

PAGE 1

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