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Section 10.2 Parametric Equations

We are familiar with sketching shapes, such as parabolas, by following this basic procedure:

Figure 10.2.1. Plotting a graph \(y=f(x)\)

The rectangular equation \(y=f(x)\) works well for some shapes like a parabola with a vertical axis of symmetry, but in the previous section we encountered several shapes that could not be sketched in this manner. (To plot an ellipse using the above procedure, we need to plot the “top” and “bottom” separately.)

In this section we introduce a new sketching procedure:

Figure 10.2.2. Plotting a curve \((x(t),y(t))\)

Here, \(x\) and \(y\) are found separately but then plotted together: for each value of the input \(t\text{,}\) we plot the output - the point \((x(t),y(t))\text{.}\)

Subsection 10.2.1 Plotting parametric curves

The procedure outlined in Figure 10.2.2 leads us to a definition.

Definition 10.2.3. Parametric Equations and Curves.

Let \(f\) and \(g\) be continuous functions on an interval \(I\text{.}\) The set of all points \(\big(x,y\big) = \big(f(t),g(t)\big)\) in the Cartesian plane, as \(t\) varies over \(I\text{,}\) is the graph of the parametric equations \(x=f(t)\) and \(y=g(t)\text{,}\) where \(t\) is the parameter. A curve is a graph along with the parametric equations that define it.

This is a formal definition of the word curve. When a curve lies in a plane (such as the Cartesian plane), it is often referred to as a plane curve. Examples will help us understand the concepts introduced in the definition.

Example 10.2.4. Plotting parametric functions.

Plot the graph of the parametric equations \(x=t^2\text{,}\) \(y=t+1\) for \(t\) in \([-2,2]\text{.}\)

Solution.

We plot the graphs of parametric equations in much the same manner as we plotted graphs of functions like \(y=f(x)\text{:}\) we make a table of values, plot points, then connect these points with a “reasonable” looking curve. Figure 10.2.5.(a) shows such a table of values; note how we have 3 columns.

The points \((x,y)\) from the table are plotted in Figure 10.2.5.(b). The points have been connected with a smooth curve. Each point has been labeled with its corresponding \(t\)-value. These values, along with the two arrows along the curve, are used to indicate the orientation of the graph. This information helps us determine the direction in which the graph is “moving.”

\(t\) \(x\) \(y\)
\(-2\) \(4\) \(-1\)
\(-1\) \(1\) \(0\)
\(0\) \(0\) \(1\)
\(1\) \(1\) \(2\)
\(2\) \(4\) \(3\)
(a)
(b)
Figure 10.2.5. A table of values of the parametric equations in Example 10.2.4 along with a sketch of their graph

We often use the letter \(t\) as the parameter as we often regard \(t\) as representing time. Certainly there are many contexts in which the parameter is not time, but it can be helpful to think in terms of time as one makes sense of parametric plots and their orientation (for instance, “At time \(t=0\) the position is \((1,2)\) and at time \(t=3\) the position is \((5,1)\text{.}\)”).

Example 10.2.6. Plotting parametric functions.

Sketch the graph of the parametric equations \(x=\cos^2(t)\text{,}\) \(y=\cos(t) +1\) for \(t\) in \([0,\pi]\text{.}\)

Solution.

We again start by making a table of values in Figure 10.2.7.(a), then plot the points \((x,y)\) on the Cartesian plane in Figure 10.2.7.(b).

\(t\) \(x\) \(y\)
\(0\) \(1\) \(2\)
\(\pi/4\) \(1/2\) \(1+\sqrt{2}/2\)
\(\pi/2\) \(0\) \(1\)
\(3\pi/4\) \(1/2\) \(1-\sqrt{2}/2\)
\(\pi\) \(1\) \(0\)
(a)
(b)
Figure 10.2.7. A table of values of the parametric equations in Example 10.2.6 along with a sketch of their graph

It is not difficult to show that the curves in Examples 10.2.4 and Example 10.2.6 are portions of the same parabola. While the parabola is the same, the curves are different. In Example 10.2.4, if we let \(t\) vary over all real numbers, we'd obtain the entire parabola. In this example, letting \(t\) vary over all real numbers would still produce the same graph; this portion of the parabola would be traced, and re-traced, infinitely many times. The orientation shown in Figure 10.2.7 shows the orientation on \([0,\pi]\text{,}\) but this orientation is reversed on \([\pi,2\pi]\text{.}\)

These examples begin to illustrate the powerful nature of parametric equations. Their graphs are far more diverse than the graphs of functions produced by “\(y=f(x)\)” functions.

Technology Note: Most graphing utilities can graph functions given in parametric form. Often the word “parametric” is abbreviated as “PAR” or “PARAM” in the options. The user usually needs to determine the graphing window (i.e, the minimum and maximum \(x\)- and \(y\)-values), along with the values of \(t\) that are to be plotted. The user is often prompted to give a \(t\) minimum, a \(t\) maximum, and a “\(t\)-step” or “\(\Delta t\text{.}\)” Graphing utilities effectively plot parametric functions just as we've shown here: they plots lots of points. A smaller \(t\)-step plots more points, making for a smoother graph (but may take longer). In Figure 10.2.5, the \(t\)-step is 1; in Figure 10.2.7, the \(t\)-step is \(\pi/4\text{.}\)

One nice feature of parametric equations is that their graphs are easy to shift. While this is not too difficult in the “\(y=f(x)\)” context, the resulting function can look rather messy. (Plus, to shift to the right by two, we replace \(x\) with \(x-2\text{,}\) which is counter-intuitive.) The following example demonstrates this.

Example 10.2.8. Shifting the graph of parametric functions.

Sketch the graph of the parametric equations \(x=t^2+t\text{,}\) \(y=t^2-t\text{.}\) Find new parametric equations that shift this graph to the right 3 places and down 2.

Solution.

The graph of the parametric equations is given in Figure 10.2.9.(a). It is a parabola with a axis of symmetry along the line \(y=x\text{;}\) the vertex is at \((0,0)\text{.}\)

In order to shift the graph to the right 3 units, we need to increase the \(x\)-value by 3 for every point. The straightforward way to accomplish this is simply to add 3 to the function defining \(x\text{:}\) \(x = t^2+t+3\text{.}\) To shift the graph down by 2 units, we wish to decrease each \(y\)-value by 2, so we subtract 2 from the function defining \(y\text{:}\) \(y = t^2-t-2\text{.}\) Thus our parametric equations for the shifted graph are \(x=t^2+t+3\text{,}\) \(y=t^2-t-2\text{.}\) This is graphed in Figure 10.2.9.(a). Notice how the vertex is now at \((3,-2)\text{.}\)

(a)
(b)
Figure 10.2.9. Illustrating how to shift graphs in Example 10.2.8

Because the \(x\)- and \(y\)-values of a graph are determined independently, the graphs of parametric functions often possess features not seen on “\(y=f(x)\)” type graphs. The next example demonstrates how such graphs can arrive at the same point more than once.

Example 10.2.10. Graphs that cross themselves.

Plot the parametric functions \(x=t^3-5t^2+3t+11\) and \(y=t^2-2t+3\) and determine the \(t\)-values where the graph crosses itself.

Solution.

Using the methods developed in this section, we again plot points and graph the parametric equations as shown in Figure 10.2.11. It appears that the graph crosses itself at the point \((2,6)\text{,}\) but we'll need to analytically determine this.

Figure 10.2.11. A graph of the parametric equations from Example 10.2.10

We are looking for two different values, say, \(s\) and \(t\text{,}\) where \(x(s) = x(t)\) and \(y(s) = y(t)\text{.}\) That is, the \(x\)-values are the same precisely when the \(y\)-values are the same. This gives us a system of 2 equations with 2 unknowns:

\begin{align*} s^3-5s^2+3s+11 \amp = t^3-5t^2+3t+11\\ s^2-2s+3 \amp = t^2-2t+3 \end{align*}

Solving this system is not trivial but involves only algebra. Using the quadratic formula, one can solve for \(t\) in the second equation and find that \(\ds t = 1\pm \sqrt{s^2-2s+1}\text{.}\) This can be substituted into the first equation, revealing that the graph crosses itself at \(t=-1\) and \(t=3\text{.}\) We confirm our result by computing \(x(-1) = x(3)=2\) and \(y(-1) = y(3) = 6\text{.}\)

Subsection 10.2.2 Converting between rectangular and parametric equations

It is sometimes useful to rewrite equations in rectangular form (i.e., \(y=f(x)\)) into parametric form, and vice-versa. Converting from rectangular to parametric can be very simple: given \(y=f(x)\text{,}\) the parametric equations \(x=t\text{,}\) \(y=f(t)\) produce the same graph. As an example, given \(y=x^2\text{,}\) the parametric equations \(x=t\text{,}\) \(y=t^2\) produce the familiar parabola. However, other parametrizations can be used. The following example demonstrates one possible alternative.

Example 10.2.12. Converting from rectangular to parametric.

Consider \(y=x^2\text{.}\) Find parametric equations \(x=f(t)\text{,}\) \(y=g(t)\) for the parabola where \(t=\frac{dy}{dx}\text{.}\) That is, \(t=a\) corresponds to the point on the graph whose tangent line has slope \(a\text{.}\)

Solution.

We start by computing \(\frac{dy}{dx}\text{:}\) \(y' = 2x\text{.}\) Thus we set \(t=2x\text{.}\) We can solve for \(x\) and find \(x= t/2\text{.}\) Knowing that \(y=x^2\text{,}\) we have \(y= t^2/4\text{.}\) Thus parametric equations for the parabola \(y=x^2\) are

\begin{equation*} x=t/2 y=t^2/4\text{.} \end{equation*}

To find the point where the tangent line has a slope of \(-2\text{,}\) we set \(t=-2\text{.}\) This gives the point \((-1, 1)\text{.}\) We can verify that the slope of the line tangent to the curve at this point indeed has a slope of \(-2\text{.}\)

We sometimes choose the parameter to accurately model physical behavior.

Example 10.2.13. Converting from rectangular to parametric.

An object is fired from a height of 0 feet and lands 6 seconds later, 192 feet away. Assuming ideal projectile motion, the height, in feet, of the object can be described by \(h(x) = -x^2/64+3x\text{,}\) where \(x\) is the distance in feet from the initial location. (Thus \(h(0) = h(192) = 0\) feet.) Find parametric equations \(x=f(t)\text{,}\) \(y=g(t)\) for the path of the projectile where \(x\) is the horizontal distance the object has traveled at time \(t\) (in seconds) and \(y\) is the height at time \(t\text{.}\)

Solution.

Physics tells us that the horizontal motion of the projectile is linear; that is, the horizontal speed of the projectile is constant. Since the object travels 192 ft in 6 s, we deduce that the object is moving horizontally at a rate of 32 fts, giving the equation \(x=32t\text{.}\) As \(y=-x^2/64+3x\text{,}\) we find \(y= -16t^2+96t\text{.}\) We can quickly verify that \(y''=-32\) ftft2, the acceleration due to gravity, and that the projectile reaches its maximum at \(t=3\text{,}\) halfway along its path.

These parametric equations make certain determinations about the object's location easy: 2 seconds into the flight the object is at the point \(\big(x(2),y(2)\big) = \big(64,128\big)\text{.}\) That is, it has traveled horizontally 64 ft and is at a height of 128 ft, as shown in Figure 10.2.14.

Figure 10.2.14. Graphing projectile motion in Example 10.2.13

It is sometimes necessary to convert given parametric equations into rectangular form. This can be decidedly more difficult, as some “simple” looking parametric equations can have very “complicated” rectangular equations. This conversion is often referred to as “eliminating the parameter,” as we are looking for a relationship between \(x\) and \(y\) that does not involve the parameter \(t\text{.}\)

Example 10.2.15. Eliminating the parameter.

Find a rectangular equation for the curve described by

\begin{equation*} x= \frac{1}{t^2+1} \text{ and } y=\frac{t^2}{t^2+1}\text{.} \end{equation*}
Solution.

There is not a set way to eliminate a parameter. One method is to solve for \(t\) in one equation and then substitute that value in the second. We use that technique here, then show a second, simpler method.

Starting with \(x= 1/(t^2+1)\text{,}\) solve for \(t\text{:}\) \(t = \pm\sqrt{1/x-1}\text{.}\) Substitute this value for \(t\) in the equation for \(y\text{:}\)

\begin{align*} y \amp = \frac{t^2}{t^2 +1}\\ \amp = \frac{1/x-1}{1/x-1+1}\\ \amp = \frac{1/x - 1}{1/x}\\ \amp = \left(\frac1x-1\right)\cdot x\\ \amp = 1-x\text{.} \end{align*}
Figure 10.2.16. Graphing parametric and rectangular equations for a graph in Example 10.2.15

Thus \(y=1-x\text{.}\) One may have recognized this earlier by manipulating the equation for \(y\text{:}\)

\begin{equation*} y = \frac{t^2}{t^2+1} = 1-\frac{1}{t^2+1} = 1-x\text{.} \end{equation*}

This is a shortcut that is very specific to this problem; sometimes shortcuts exist and are worth looking for.

We should be careful to limit the domain of the function \(y=1-x\text{.}\) The parametric equations limit \(x\) to values in \((0,1]\text{,}\) thus to produce the same graph we should limit the domain of \(y=1-x\) to the same.

The graphs of these functions is given in Figure 10.2.16. The portion of the graph defined by the parametric equations is given in a thick line; the graph defined by \(y=1-x\) with unrestricted domain is given in a thin line.

Example 10.2.17. Eliminating the parameter.

Eliminate the parameter in \(x=4\cos(t) +3\text{,}\) \(y= 2\sin(t) +1\)

Solution.

We should not try to solve for \(t\) in this situation as the resulting algebra/trig would be messy. Rather, we solve for \(\cos(t)\) and \(\sin(t)\) in each equation, respectively. This gives

\begin{equation*} \cos(t) = \frac{x-3}{4} \text{ and } \sin(t) =\frac{y-1}{2}\text{.} \end{equation*}

The Pythagorean Theorem gives \(\cos^2(t) +\sin^2(t) =1\text{,}\) so:

\begin{align*} \cos^2(t) +\sin^2(t) \amp =1\\ \left(\frac{x-3}{4}\right)^2 +\left(\frac{y-1}{2}\right)^2 \amp =1\\ \frac{(x-3)^2}{16}+\frac{(y-1)^2}{4} \amp =1 \end{align*}
Figure 10.2.18. Graphing the parametric equations \(x=4\cos(t) +3\text{,}\) \(y=2\sin(t) +1\) in Example 10.2.17

This final equation should look familiar — it is the equation of an ellipse! Figure 10.2.18 plots the parametric equations, demonstrating that the graph is indeed of an ellipse with a horizontal major axis and center at \((3,1)\text{.}\)

The Pythagorean Theorem can also be used to identify parametric equations for hyperbolas. We give the parametric equations for ellipses and hyperbolas in the following Key Idea.

Key Idea 10.2.19. Parametric Equations of Ellipses and Hyperbolas.

  • The parametric equations

    \begin{equation*} x=a\cos(t) +h, y=b\sin(t) +k \end{equation*}

    define an ellipse with horizontal axis of length \(2a\) and vertical axis of length \(2b\text{,}\) centered at \((h,k)\text{.}\)

  • The parametric equations

    \begin{equation*} x= a\tan(t) +h, y=\pm b\sec(t) +k \end{equation*}

    define a hyperbola with vertical transverse axis centered at \((h,k)\text{,}\) and

    \begin{equation*} x=\pm a\sec(t) +h, y=b\tan(t) + k \end{equation*}

    defines a hyperbola with horizontal transverse axis. Each has asymptotes at \(y=\pm b/a(x-h)+k\text{.}\)

Subsection 10.2.3 Special Curves

Figure 10.2.20 gives a small gallery of “interesting” and “famous” curves along with parametric equations that produce them. Interested readers can begin learning more about these curves through internet searches.

One might note a feature shared by two of these graphs: “sharp corners,” or cusps. We have seen graphs with cusps before and determined that such functions are not differentiable at these points. This leads us to a definition.

(a) Astroid where \(x=\cos^3(t)\) and \(y=\sin^3(t)\)
(b) Rose Curve where \(x=\cos(t)\sin(4t)\) and \(y=\sin(t)\sin(4t)\)
(c) Hypotrochoid where \(x=2\cos(t)+5\cos(2t/3)\) and \(y=2\sin(t)-5\sin(2t/3)\)
(d) Epicycloid where \(x=4\cos(t)-\cos(4t)\) and \(y=4\sin(t)-\sin(4t)\)
Figure 10.2.20. A gallery of interesting planar curves

Definition 10.2.21. Smooth.

A curve \(C\) defined by \(x=f(t)\text{,}\) \(y=g(t)\) is smooth on an interval \(I\) if \(\fp\) and \(g'\) are continuous on \(I\) and not simultaneously 0 (except possibly at the endpoints of \(I\)). A curve is piecewise smooth on \(I\) if \(I\) can be partitioned into subintervals where \(C\) is smooth on each subinterval.

Consider the astroid, given by \(x=\cos^3(t)\text{,}\) \(y=\sin^3(t)\text{.}\) Taking derivatives, we have:

\begin{equation*} x' = -3\cos^2(t) \sin(t) \text{ and } y' = 3\sin^2(t) \cos(t)\text{.} \end{equation*}

It is clear that each is 0 when \(t=0,\, \pi/2,\, \pi,\ldots\text{.}\) Thus the astroid is not smooth at these points, corresponding to the cusps seen in the figure.

We demonstrate this once more.

Example 10.2.22. Determine where a curve is not smooth.

Let a curve \(C\) be defined by the parametric equations \(x=t^3-12t+17\) and \(y=t^2-4t+8\text{.}\) Determine the points, if any, where it is not smooth.

Solution.

We begin by taking derivatives.

\begin{equation*} x' = 3t^2-12, y' = 2t-4\text{.} \end{equation*}

We set each equal to 0:

\begin{align*} x' \amp = 0 \Rightarrow 3t^2-12=0 \Rightarrow t=\pm 2\\ y' \amp = 0 \Rightarrow 2t-4 = 0 \Rightarrow t=2 \end{align*}

We see at \(t=2\) both \(x'\) and \(y'\) are 0; thus \(C\) is not smooth at \(t=2\text{,}\) corresponding to the point \((1,4)\text{.}\) The curve is graphed in Figure 10.2.23, illustrating the cusp at \((1,4)\text{.}\)

Figure 10.2.23. Graphing the curve in Example 10.2.22; note it is not smooth at \((1,4)\)

If a curve is not smooth at \(t=t_0\text{,}\) it means that \(x'(t_0)=y'(t_0)=0\) as defined. This, in turn, means that rate of change of \(x\) (and \(y\)) is 0; that is, at that instant, neither \(x\) nor \(y\) is changing. If the parametric equations describe the path of some object, this means the object is at rest at \(t_0\text{.}\) An object at rest can make a “sharp” change in direction, whereas moving objects tend to change direction in a “smooth” fashion.

One should be careful to note that a “sharp corner” does not have to occur when a curve is not smooth. For instance, one can verify that \(x=t^3\text{,}\) \(y=t^6\) produce the familiar \(y=x^2\) parabola. However, in this parametrization, the curve is not smooth. A particle traveling along the parabola according to the given parametric equations comes to rest at \(t=0\text{,}\) though no sharp point is created.

Our previous experience with cusps taught us that a function was not differentiable at a cusp. This can lead us to wonder about derivatives in the context of parametric equations and the application of other calculus concepts. Given a curve defined parametrically, how do we find the slopes of tangent lines? Can we determine concavity? We explore these concepts and more in the next section.

Exercises 10.2.4 Exercises

Terms and Concepts

1.

What is the difference between degenerate and nondegenerate conics?

Answer.

When defining the conics as the intersections of a plane and a double napped cone, degenerate conics are created when the plane intersects the tips of the cones (usually taken as the origin). Nondegenerate conics are formed when this plane does not contain the origin.

2.

Use your own words to explain what the eccentricity of an ellipse measures.

Answer.

Answers will vary.

3.

What has the largest eccentricity: an ellipse or a hyperbola?

Answer.

Hyperbola

4.

Explain why the following is true: “If the coefficient of the \(x^2\) term in the equation of an ellipse in standard form is smaller than the coefficient of the \(y^2\) term, then the ellipse has a horizontal major axis.”

Answer.

With the equation \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\text{,}\) the ellipse has a horizontal major axis if \(a \gt b\text{.}\) But the coefficient of the \(x^2\) term is \(1/a^2\) (not \(a^2\)), so if \(1/a^2\lt 1/b^2\text{,}\) then \(a \gt b\) and the major axis is horizontal.

5.

Explain how one can quickly look at the equation of a hyperbola in standard form and determine whether the transverse axis is horizontal or vertical.

Answer.

With a horizontal transverse axis, the \(x^2\) term has a positive coefficient; with a vertical transverse axis, the \(y^2\) term has a positive coefficient.

6.

Fill in the blank: It can be said that ellipses and hyperbolas share the same reflective property: “A ray emanating from one focus will reflect off the conic along a that contains the other focus.”

Answer.

line

Problems

Exercise Group.

In the following exercises, find the equation of the parabola defined by the given information. Sketch the parabola.

7.

Focus: \((3,2)\text{;}\) directrix: \(y=1\)

Answer.

\(y=\frac12(x-3)^2+\frac32\)

8.

Focus: \((-1,-4)\text{;}\) directrix: \(y=2\)

Answer.

\(y=\frac{-1}{12}(x+1)^2-1\)

9.

Focus: \((1,5)\text{;}\) directrix: \(x=3\)

Answer.

\(x=-\frac14(y-5)^2+2\)

10.

Focus: \((1/4,0)\text{;}\) directrix: \(x=-1/4\)

Answer.

\(x=y^2\)

11.

Focus: \((1,1)\text{;}\) vertex: \((1,2)\)

Answer.

\(y=-\frac14(x-1)^2+2\)

12.

Focus: \((-3,0)\text{;}\) vertex: \((0,0)\)

Answer.

\(x=-\frac1{12}y^2\)

13.

Vertex: \((0,0)\text{;}\) directrix: \(y=-1/16\)

Answer.

\(y=4x^2\)

14.

Vertex: \((2,3)\text{;}\) directrix: \(x=4\)

Answer.

\(x=-\frac18(y-3)^2+2\)

Exercise Group.

In the following exercises, the equation of a parabola and a point on its graph are given. Find the focus and directrix of the parabola, and verify that the given point is equidistant from the focus and directrix.

15.

\(y=\frac14x^2\text{,}\) \(P=(2,1)\)

Answer.

focus: \((0,1)\text{;}\) directrix: \(y=-1\text{.}\) The point \(P\) is 2 units from each.

16.

\(x=\frac18(y-2)^2+3\text{,}\) \(P=(11,10)\)

Answer.

focus: \((5,2)\text{;}\) directrix: \(x=1\text{.}\) The point \(P\) is 10 units from each.

Exercise Group.

In the following exercises, sketch the ellipse defined by the given equation. Label the center, foci and vertices.

17.

\(\ds \frac{(x-1)^2}{3}+\frac{(y-2)^2}{5}=1\)

18.

\(\ds \frac{1}{25}x^2+\frac{1}{9}(y+3)^2=1\)

Exercise Group.

In the following exercises, find the equation of the ellipse shown in the graph. Give the location of the foci and the eccentricity of the ellipse.

19.
Answer.

\(\frac{(x+1)^2}{9}+\frac{(y-2)^2}{4}=1\text{;}\) foci at \((-1\pm\sqrt{5},2)\text{;}\) \(e=\sqrt{5}/3\)

20.
Answer.

\(\frac{(x-1)^2}{1/4}+\frac{y^2}{9}=1\text{;}\) foci at \((1,\pm \sqrt{8.75})\text{;}\) \(e=\sqrt{8.75}/3\approx 0.99\)

Exercise Group.

In the following exercises, find the equation of the ellipse defined by the given information. Sketch the elllipse.

21.

Foci: \((\pm 2,0)\text{;}\) vertices: \((\pm 3,0)\)

Answer.

\(\frac{x^2}{9}+\frac{y^2}{5}=1\)

22.

Foci: \((-1,3)\) and \((5,3)\text{;}\) vertices: \((-3,3)\) and \((7,3)\)

Answer.

\(\frac{(x-2)^2}{25}+\frac{(y-3)^2}{16}=1\)

23.

Foci: \((2,\pm 2)\text{;}\) vertices: \((2,\pm 7)\)

Answer.

\(\frac{(x-2)^2}{45}+\frac{y^2}{49}=1\)

24.

Focus: \((-1,5)\text{;}\) vertex: \((-1,-4)\text{;}\) center: \((-1,1)\)

Answer.

\(\frac{(x+1)^2}{9}+\frac{(y-1)^2}{25}=1\)

Exercise Group.

In the following exercises, write the equation of the given ellipse in standard form.

25.

\(x^2-2x+2y^2-8y=-7\)

Answer.

\(\frac{(x-1)^2}{2}+(y-2)^2=1\)

26.

\(5x^2+3y^2=15\)

Answer.

\(\frac{x^2}{3}+\frac{y^2}{5}=1\)

27.

\(3x^2+2y^2-12y+6=0\)

Answer.

\(\frac{x^2}{4}+\frac{(y-3)^2}{6}=1\)

28.

\(x^2+y^2-4x-4y+4=0\)

Answer.

\(\frac{(x-2)^2}{4}+\frac{(y-2)^2}{4}=1\)

Exercise Group.

In the following exercises, find the equation of the hyperbola shown in the graph.

29.
Answer.

\(x^2-\frac{y^2}{3}=1\)

30.
Answer.

\(y^2-\frac{x^2}{24}=1\)

31.
Answer.

\(\frac{(y-3)^2}{4}-\frac{(x-1)^2}{9}=1\)

32.
Answer.

\(\frac{(x-1)^2}{9}-\frac{(y-3)^2}{4}=1\)

Exercise Group.

In the following exercises, sketch the hyperbola defined by the given equation. Label the center and foci.

33.

\(\ds \frac{(x-1)^2}{16}-\frac{(y+2)^2}{9}=1\)

34.

\(\ds (y-4)^2-\frac{(x+1)^2}{25}=1\)

Exercise Group.

In the following exercises, find the equation of the hyperbola defined by the given information. Sketch the hyperbola.

35.

Foci: \((\pm 3,0)\text{;}\) vertices: \((\pm 2, 0)\)

Answer.

\(\frac{x^2}{4}-\frac{y^2}{5}=1\)

36.

Foci: \((0,\pm 3)\text{;}\) vertices: \((0,\pm 2)\)

Answer.

\(\frac{y^2}{4}-\frac{x^2}{5}=1\)

37.

Foci: \((-2,3)\) and \((8,3)\text{;}\) vertices: \((-1,3)\) and \((7,3)\)

Answer.

\(\frac{(x-3)^2}{16}-\frac{(y-3)^2}{9}=1\)

38.

Foci: \((3,-2)\) and \((3,8)\text{;}\) vertices: \((3,0)\) and \((3,6)\)

Answer.

\(\frac{(y-3)^2}{9}-\frac{(x-3)^2}{16}=1\)

Exercise Group.

In the following exercises, write the equation of the hyperbola in standard form.

39.

\(3x^2-4y^2=12\)

Answer.

\(\frac{x^2}{4}-\frac{y^2}{3}=1\)

40.

\(3x^2-y^2+2y=10\)

Answer.

\(\frac{x^2}{3}-\frac{(y-1)^2}{9}=1\)

41.

\(x^2-10y^2+40y=30\)

Answer.

\((y-2)^2-\frac{x^2}{10}=1\)

42.

\((4y-x)(4y+x)=4\)

Answer.

\(4y^2-\frac{x^2}{4}=1\)

43.

Consider the ellipse given by \(\ds \frac{(x-1)^2}{4}+\frac{(y-3)^2}{12}=1\text{.}\)

  1. Verify that the foci are located at \((1,3\pm 2\sqrt{2})\text{.}\)

  2. The points \(P_1 = (2,6)\) and \(P_2 = (1+\sqrt{2},3+\sqrt{6}) \approx (2.414,5.449)\) lie on the ellipse. Verify that the sum of distances from each point to the foci is the same.

Answer.
  1. \(c=\sqrt{12-4} = 2\sqrt{2}\text{.}\)

  2. The sum of distances for each point is \(2\sqrt{12}\approx 6.9282\text{.}\)

44.

Johannes Kepler discovered that the planets of our solar system have elliptical orbits with the Sun at one focus. The Earth's elliptical orbit is used as a standard unit of distance; the distance from the center of Earth's elliptical orbit to one vertex is 1 Astronomical Unit, or A.U.

The following table gives information about the orbits of three planets.

Planet Distance from
center to vertex
Orbit
eccentricity
Mercury \(0.387\) A.U. \(0.2056\)
Earth 1 A.U. \(0.0167\)
Mars \(1.524\) A.U. \(0.0934\)
  1. In an ellipse, knowing \(c^2=a^2-b^2\) and \(e=c/a\) allows us to find \(b\) in terms of \(a\) and \(e\text{.}\) Show \(b=a\sqrt{1-e^2}\text{.}\)

  2. For each planet, find equations of their elliptical orbit of the form \(\ds\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{.}\) (This places the center at \((0,0)\text{,}\) but the Sun is in a different location for each planet.)

  3. Shift the equations so that the Sun lies at the origin. Plot the three elliptical orbits.

Answer.
  1. Solve for \(c\) in \(e=c/a\text{:}\) \(c=ae\text{.}\) Thus \(a^2e^2=a^2-b^2\text{,}\) and \(b^2=a^2-a^2e^2\text{.}\) The result follows.

  2. Mercury: \(x^2/(0.387)^2 + y^2/(0.3787)^2=1\) Earth: \(x^2+y^2/(0.99986)^2 = 1\) Mars: \(x^2/(1.524)^2+y^2/(1.517)^2=1\)

  3. Mercury: \((x-0.08)^2/(0.387)^2 + y^2/(0.3787)^2=1\) Earth: \((x-0.0167)^2+y^2/(0.99986)^2 = 1\) Mars: \((x-0.1423)^2/(1.524)^2+y^2/(1.517)^2=1\)

45.

A loud sound is recorded at three stations that lie on a line as shown in the figure below. Station \(A\) recorded the sound 1 second after Station \(B\text{,}\) and Station \(C\) recorded the sound 3 seconds after \(B\text{.}\) Using the speed of sound as 340m/s, determine the location of the sound's origination.

Answer.

The sound originated from a point approximately 31m to the left of \(B\) and 1340m above it.