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The Ellipse

Write equations of ellipses in standard form.

The Ellipse

  • Complete the square of a binomial expression. (IA 9.2.1)
  • Graph a circle. (IA 11.1.4)

But what happens if we have to solve an equation where the trinomial is not a perfect square?

For example, \({x}^{2}+4x+5=2\) ? For these types of equations, we can use a process called completing the square.

Recall \({(x+1)}^{2}=(x+1)(x+1)={x}^{2}+x+x+1={x}^{2}+2+1\) .

We can use the Binomial Squares Pattern to make a perfect square.

Example

Try it.

Complete the square for \({x}^{2}+6x\) to make it a perfect square.

Solution
Since there is a plus sign between the two terms, we will use the (a + b)2 pattern \({a}^{2}+2ab+{b}^{2}={(a+b)}^{2}\) . \({x}^{2}+6x\)
We ultimately need to find the last term of this trinomial that will make it a perfect square trinomial. To do that we will need to find b. But first we start with determining a. Notice that the first term of x2 + 6x is a square, x2. This tells us that a = x. \({x}^{2}+2\cdot x\cdot b+{b}^{2}\)
What number, b, when multiplied with 2x, gives 6x? It would have to be 3, which is (½)(6).
So b = 3.		 \({x}^{2}+2\cdot 3\cdot x+_\)
Now to complete the perfect square trinomial, we will find the last term by squaring b, which is 32 = 9. \({x}^{2}+6x+9\)
We can now factor. \({(x+3)}^{2}\)

So, we found that adding 9 to x2 + 6x completes the square, and we write it as (x + 3)2.

Determine what number would have to be added to the given terms to create a perfect square trinomial. Then rewrite as a binomial squared.

Try it.

\({x}^{2}+12x\)

Try it.

\({x}^{2}+5x\)

Try it.

\({x}^{2}-\frac{1}{2}x\)

Try it.

\({x}^{2}+\frac{3}{2}x\)

Condensed — the full section is in OpenStax Precalculus 2e.

Writing Equations of Ellipses in Standard Form

A conic section, or conic, is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape, as shown in .

Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter, we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the equation for the ellipse. An ellipse is the set of all points \((x,y)\) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci).

We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. See .

Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis. Each endpoint of the major axis is the vertex of the ellipse (plural: vertices), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci. See .

In this section, we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the x- and y-axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane.

To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs.

Condensed — the full section is in OpenStax Precalculus 2e.

Graphing Ellipses Centered at the Origin

Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\ a>b\) for horizontal ellipses and \(\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,\ a>b\) for vertical ellipses.

Example

Try it.

Graph the ellipse given by the equation, \(\frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1.\) Identify and label the center, vertices, co-vertices, and foci.

Solution

First, we determine the position of the major axis. Because \(25>9,\) the major axis is on the y-axis. Therefore, the equation is in the form \(\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,\) where \({b}^{2}=9\) and \({a}^{2}=25.\) It follows that:

  • the center of the ellipse is \((0,0)\)
  • the coordinates of the vertices are \((0,\pm a)=(0,\pm \sqrt{25})=(0,\pm 5)\)
  • the coordinates of the co-vertices are \((\pm b,0)=(\pm \sqrt{9},0)=(\pm 3,0)\)
  • the coordinates of the foci are \((0,\pm c),\) where \({c}^{2}={a}^{2}-{b}^{2}\) Solving for \(c,\) we have:

\[\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}} \\ =\pm \sqrt{25-9} \\ =\pm \sqrt{16} \\ =\pm 4\end{array}\]

Therefore, the coordinates of the foci are \((0,\pm 4).\)

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See .

Condensed — the full section is in OpenStax Precalculus 2e.

Graphing Ellipses Not Centered at the Origin

When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, \((h,k),\) we use the standard forms \(\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1,\ a>b\) for horizontal ellipses and \(\frac{{(x-h)}^{2}}{{b}^{2}}+\frac{{(y-k)}^{2}}{{a}^{2}}=1,\ a>b\) for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.

Condensed — the full section is in OpenStax Precalculus 2e.

Solving Applied Problems Involving Ellipses

Many real-world situations can be represented by ellipses, including orbits of planets, satellites, moons and comets, and shapes of boat keels, rudders, and some airplane wings. A medical device called a lithotripter uses elliptical reflectors to break up kidney stones by generating sound waves. Some buildings, called whispering chambers, are designed with elliptical domes so that a person whispering at one focus can easily be heard by someone standing at the other focus. This occurs because of the acoustic properties of an ellipse. When a sound wave originates at one focus of a whispering chamber, the sound wave will be reflected off the elliptical dome and back to the other focus. See . In the whisper chamber at the Museum of Science and Industry in Chicago, two people standing at the foci—about 43 feet apart—can hear each other whisper. When these chambers are placed in unexpected places, such as the ones inside Bush International Airport in Houston and Grand Central Terminal in New York City, they can induce surprised reactions among travelers.

Condensed — the full section is in OpenStax Precalculus 2e.

Key Equations

Horizontal ellipse, center at origin \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\ a>b\)
Vertical ellipse, center at origin \(\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,\ a>b\)
Horizontal ellipse, center \((h,k)\) \(\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1,\ a>b\)
Vertical ellipse, center \((h,k)\) \(\frac{{(x-h)}^{2}}{{b}^{2}}+\frac{{(y-k)}^{2}}{{a}^{2}}=1,\ a>b\)

Key Concepts

  • An ellipse is the set of all points \((x,y)\) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci).
  • When given the coordinates of the foci and vertices of an ellipse, we can write the equation of the ellipse in standard form. See and .
  • When given an equation for an ellipse centered at the origin in standard form, we can identify its vertices, co-vertices, foci, and the lengths and positions of the major and minor axes in order to graph the ellipse. See and .
  • When given the equation for an ellipse centered at some point other than the origin, we can identify its key features and graph the ellipse. See and .
  • Real-world situations can be modeled using the standard equations of ellipses and then evaluated to find key features, such as lengths of axes and distance between foci. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Use the square root property to solve \({(x+3)}^{2}=5\) . Use the steps below to guide your work.

    1. ⓐ Take a square root of both sides. Remember to have ± on the right side.
    2. ⓑ Solve for x.

  2. Use the square root property to solve \({x}^{2}+6x+9=7\). Use the steps below to guide your work.

    1. ⓐ \({x}^{2}+6x+9\) is a perfect square trinomial. Factor it into \({(_+_)}^{2}\) .
    2. ⓑ Take a square root of both sides. Remember to have ± on the right side.
    3. ⓒ Solve for x.

  3. Complete the square for \({x}^{2}+6x\) to make it a perfect square.

    Cevabı açıkla.
    Since there is a plus sign between the two terms, we will use the (a + b)2 pattern \({a}^{2}+2ab+{b}^{2}={(a+b)}^{2}\) . \({x}^{2}+6x\)
    We ultimately need to find the last term of this trinomial that will make it a perfect square trinomial. To do that we will need to find b. But first we start with determining a. Notice that the first term of x2 + 6x is a square, x2. This tells us that a = x. \({x}^{2}+2\cdot x\cdot b+{b}^{2}\)
    What number, b, when multiplied with 2x, gives 6x? It would have to be 3, which is (½)(6).
    So b = 3.    		 \({x}^{2}+2\cdot 3\cdot x+_\)
    Now to complete the perfect square trinomial, we will find the last term by squaring b, which is 32 = 9. \({x}^{2}+6x+9\)
    We can now factor. \({(x+3)}^{2}\)

    So, we found that adding 9 to x2 + 6x completes the square, and we write it as (x + 3)2.

  4. \({x}^{2}+12x\)

  5. \({x}^{2}+5x\)

  6. \({x}^{2}-\frac{1}{2}x\)

  7. \({x}^{2}+\frac{3}{2}x\)

  8. Write the standard (graphing) form of the equation of the circle with radius 2 and center (-1, 3).

    Cevabı açıkla.

    Use the standard, (graphing) form of the equation of a circle. \({(x-h)}^{2}+{(y-k)}^{2}={r}^{2}\)
    Substitute in the values (h, k)=(1, -3), where h=1, k=-3. \((x-{(-1))}^{2}+(y-{(3))}^{2}={2}^{2}\)
    Simplify. \({(x+1)}^{2}+{(y-3)}^{2}=4\)

  9. Find the center and radius, then graph the circle: \({(x+2)}^{2}+{(y-1)}^{2}=9\)

    Cevabı açıkla.

    Use the standard (graphing) form of the equation of a circle. \({(x-h)}^{2}+{(y-k)}^{2}={r}^{2}\)
    Identify the center, (h,k) and radius, r. \((x-{(-2))}^{2}+{(y-1)}^{2}={3}^{2}\)
    Graph the circle. Center (-2, 1), r=3

  10. Find the center and radius, then graph the circle: \({x}^{2}+{y}^{2}-4x-6y+4=0\)

    Cevabı açıkla.

    We need to rewrite this general form into standard (graphing) form in order to find the center and radius.

    Group the x-terms and y-terms. Collect the constants on the right side. \({x}^{2}-4x+{y}^{2}-6y=-4\)
    Complete the squares. \({x}^{2}-4x+4+{y}^{2}-6y+9=-4+4+9\)
    Rewrite as binomial squares. \({(x-2)}^{2}+{(y-3)}^{2}=9\)
    Identify the center and radius.Center (2, 3), r=3
    Graph the circle.

  11. Write the standard (graphing) form of the equation of the circle with radius 4 and center (2,–5).

  12. Find the center and radius, then graph the circle: \({(x-3)}^{2}+{(y+1)}^{2}=4\)

  13. Find the center and radius, then graph the circle: \({x}^{2}+{y}^{2}-6x-8y+9=0\)

  14. What is the standard form equation of the ellipse that has vertices \((\pm 8,0)\) and foci \((\pm 5,0)?\)

    Cevabı açıkla.

    The foci are on the x-axis, so the major axis is the x-axis. Thus, the equation will have the form

    \[\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\]

    The vertices are \((\pm 8,0),\) so \(a=8\) and \({a}^{2}=64.\)

    The foci are \((\pm 5,0),\) so \(c=5\) and \({c}^{2}=25.\)

    We know that the vertices and foci are related by the equation \({c}^{2}={a}^{2}-{b}^{2}.\) Solving for \({b}^{2},\) we have:

    \[\begin{array}{lllll}{c}^{2}={a}^{2}-{b}^{2} & \\ 25=64-{b}^{2}\begin{array}{llll} & & & \end{array} & \text{Substitute for }{c}^{2}\ \text{and }{a}^{2}. \\ {b}^{2}=39 & \text{Solve for }{b}^{2}.\end{array}\]

    Now we need only substitute \({a}^{2}=64\) and \({b}^{2}=39\) into the standard form of the equation. The equation of the ellipse is \(\frac{{x}^{2}}{64}+\frac{{y}^{2}}{39}=1.\)

  15. What is the standard form equation of the ellipse that has vertices \((0,\pm 4)\) and foci \((0,\pm \sqrt{15})?\)

    Cevabı açıkla.

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

  16. What is the standard form equation of the ellipse that has vertices \((-2,-8)\) and \((-2,\text{2})\)

    and foci \((-2,-7)\) and \((-2,\text{1})?\)

    Cevabı açıkla.

    The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form

    \[\frac{{(x-h)}^{2}}{{b}^{2}}+\frac{{(y-k)}^{2}}{{a}^{2}}=1\]

    First, we identify the center, \((h,k).\) The center is halfway between the vertices, \((-2,-8)\) and \((-2,\text{2}).\) Applying the midpoint formula, we have:

    \[\begin{array}{l}(h,k)=(\frac{-2+(-2)}{2},\frac{-8+2}{2}) \\ =(-2,-3)\end{array}\]

    Next, we find \({a}^{2}.\) The length of the major axis, \(2a,\) is bounded by the vertices. We solve for \(a\) by finding the distance between the y-coordinates of the vertices.

    \[\begin{array}{l}2a=2-(-8) \\ 2a=10 \\ a=5\end{array}\]

    So \({a}^{2}=25.\)

    Now we find \({c}^{2}.\) The foci are given by \((h,k\pm c).\) So, \((h,k-c)=(-2,-7)\) and \((h,k+c)=(-2,\text{1}).\) We substitute \(k=-3\) using either of these points to solve for \(c.\)

    \[\begin{array}{l}k+c=1 \\ -3+c=1 \\ c=4\end{array}\]

    So \({c}^{2}=16.\)

    Next, we solve for \({b}^{2}\) using the equation \({c}^{2}={a}^{2}-{b}^{2}.\)

    \[\begin{array}{l}{c}^{2}={a}^{2}-{b}^{2} \\ 16=25-{b}^{2} \\ {b}^{2}=9\end{array}\]

    Finally, we substitute the values found for \(h,k,{a}^{2},\) and \({b}^{2}\) into the standard form equation for an ellipse:

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

  17. What is the standard form equation of the ellipse that has vertices \((-3,3)\) and \((5,3)\) and foci \((1-2\sqrt{3},3)\) and \((1+2\sqrt{3},3)?\)

    Cevabı açıkla.

    \(\frac{{(x-1)}^{2}}{16}+\frac{{(y-3)}^{2}}{4}=1\)

  18. Graph the ellipse given by the equation, \(\frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1.\) Identify and label the center, vertices, co-vertices, and foci.

    Cevabı açıkla.

    First, we determine the position of the major axis. Because \(25>9,\) the major axis is on the y-axis. Therefore, the equation is in the form \(\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,\) where \({b}^{2}=9\) and \({a}^{2}=25.\) It follows that:

    • the center of the ellipse is \((0,0)\)
    • the coordinates of the vertices are \((0,\pm a)=(0,\pm \sqrt{25})=(0,\pm 5)\)
    • the coordinates of the co-vertices are \((\pm b,0)=(\pm \sqrt{9},0)=(\pm 3,0)\)
    • the coordinates of the foci are \((0,\pm c),\) where \({c}^{2}={a}^{2}-{b}^{2}\) Solving for \(c,\) we have:

    \[\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}} \\ =\pm \sqrt{25-9} \\ =\pm \sqrt{16} \\ =\pm 4\end{array}\]

    Therefore, the coordinates of the foci are \((0,\pm 4).\)

    Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See .

  19. Graph the ellipse given by the equation \(\frac{{x}^{2}}{36}+\frac{{y}^{2}}{4}=1.\) Identify and label the center, vertices, co-vertices, and foci.

    Cevabı açıkla.

    center: \((0,0);\) vertices: \((\pm 6,0);\) co-vertices: \((0,\pm 2);\) foci: \((\pm 4\sqrt{2},0)\)

  20. Graph the ellipse given by the equation \(4{x}^{2}+25{y}^{2}=100.\) Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

    Cevabı açıkla.

    First, use algebra to rewrite the equation in standard form.

    \[\begin{array}{l}4{x}^{2}+25{y}^{2}=100 \\ \frac{4{x}^{2}}{100}+\frac{25{y}^{2}}{100}=\frac{100}{100} \\ \frac{{x}^{2}}{25}+\frac{{y}^{2}}{4}=1\end{array}\]

    Next, we determine the position of the major axis. Because \(25>4,\) the major axis is on the x-axis. Therefore, the equation is in the form \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\) where \({a}^{2}=25\) and \({b}^{2}=4.\) It follows that:

    • the center of the ellipse is \((0,0)\)
    • the coordinates of the vertices are \((\pm a,0)=(\pm \sqrt{25},0)=(\pm 5,0)\)
    • the coordinates of the co-vertices are \((0,\pm b)=(0,\pm \sqrt{4})=(0,\pm 2)\)
    • the coordinates of the foci are \((\pm c,0),\) where \({c}^{2}={a}^{2}-{b}^{2}.\) Solving for \(c,\) we have:

    \[\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}} \\ =\pm \sqrt{25-4} \\ =\pm \sqrt{21}\end{array}\]

    Therefore the coordinates of the foci are \((\pm \sqrt{21},0).\)

    Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

  21. Graph the ellipse given by the equation \(49{x}^{2}+16{y}^{2}=784.\) Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

    Cevabı açıkla.

    Standard form: \(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{49}=1;\) center: \((0,0);\) vertices: \((0,\pm 7);\) co-vertices: \((\pm 4,0);\) foci: \((0,\pm \sqrt{33})\)

  22. Graph the ellipse given by the equation, \(\frac{{(x+2)}^{2}}{4}+\frac{{(y-5)}^{2}}{9}=1.\) Identify and label the center, vertices, co-vertices, and foci.

    Cevabı açıkla.

    First, we determine the position of the major axis. Because \(9>4,\) the major axis is parallel to the y-axis. Therefore, the equation is in the form \(\frac{{(x-h)}^{2}}{{b}^{2}}+\frac{{(y-k)}^{2}}{{a}^{2}}=1,\) where \({b}^{2}=4\) and \({a}^{2}=9.\) It follows that:

    • the center of the ellipse is \((h,k)=(-2,\text{5})\)
    • the coordinates of the vertices are \((h,k\pm a)=(-2,5\pm \sqrt{9})=(-2,5\pm 3),\) or \((-2,\text{2})\) and \((-2,\text{8})\)
    • the coordinates of the co-vertices are \((h\pm b,k)=(-2\pm \sqrt{4},5)=(-2\pm 2,5),\) or \((-4,5)\) and \((0,\text{5})\)
    • the coordinates of the foci are \((h,k\pm c),\) where \({c}^{2}={a}^{2}-{b}^{2}.\) Solving for \(c,\) we have:
    \[\begin{array}{l}\begin{array}{l} \\ c=\pm \sqrt{{a}^{2}-{b}^{2}}\end{array} \\ =\pm \sqrt{9-4} \\ =\pm \sqrt{5}\end{array}\]

    Therefore, the coordinates of the foci are \((-2,\text{5}-\sqrt{5})\) and \((-2,\text{5+}\sqrt{5}).\)

    Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

  23. Graph the ellipse given by the equation \(\frac{{(x-4)}^{2}}{36}+\frac{{(y-2)}^{2}}{20}=1.\) Identify and label the center, vertices, co-vertices, and foci.

    Cevabı açıkla.

    Center: \((4,2);\) vertices: \((-2,2)\) and \((10,2);\) co-vertices: \((4,2-2\sqrt{5})\) and \((4,2+2\sqrt{5});\) foci: \((0,2)\) and \((8,2)\)

  24. Graph the ellipse given by the equation \(4{x}^{2}+9{y}^{2}-40x+36y+100=0.\) Identify and label the center, vertices, co-vertices, and foci.

    Cevabı açıkla.

    We must begin by rewriting the equation in standard form.

    \[4{x}^{2}+9{y}^{2}-40x+36y+100=0\]

    Group terms that contain the same variable, and move the constant to the opposite side of the equation.

    \[(4{x}^{2}-40x)+(9{y}^{2}+36y)=-100\]

    Factor out the coefficients of the squared terms.

    \[4({x}^{2}-10x)+9({y}^{2}+4y)=-100\]

    Complete the square twice. Remember to balance the equation by adding the same constants to each side.

    \[4({x}^{2}-10x+25)+9({y}^{2}+4y+4)=-100+100+36\]

    Rewrite as perfect squares.

    \[4{(x-5)}^{2}+9{(y+2)}^{2}=36\]

    Divide both sides by the constant term to place the equation in standard form.

    \[\frac{{(x-5)}^{2}}{9}+\frac{{(y+2)}^{2}}{4}=1\]

    Now that the equation is in standard form, we can determine the position of the major axis. Because \(9>4,\) the major axis is parallel to the x-axis. Therefore, the equation is in the form \(\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1,\) where \({a}^{2}=9\) and \({b}^{2}=4.\) It follows that:

    • the center of the ellipse is \((h,k)=(5,-2)\)
    • the coordinates of the vertices are \((h\pm a,k)=(5\pm \sqrt{9},-2)=(5\pm 3,-2),\) or \((2,-2)\) and \((8,-2)\)
    • the coordinates of the co-vertices are \((h,k\pm b)=(\text{5},-2\pm \sqrt{4})=(\text{5},-2\pm 2),\) or \((5,-4)\) and \((5,\text{0})\)
    • the coordinates of the foci are \((h\pm c,k),\) where \({c}^{2}={a}^{2}-{b}^{2}.\) Solving for \(c,\) we have:
    \[\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}} \\ =\pm \sqrt{9-4} \\ =\pm \sqrt{5}\end{array}\]

    Therefore, the coordinates of the foci are \((\text{5}-\sqrt{5},-2)\) and \((\text{5+}\sqrt{5},-2).\)

    Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse as shown in .

  25. Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse.

    \[4{x}^{2}+{y}^{2}-24x+2y+21=0\]
    Cevabı açıkla.

    \(\frac{{(x-3)}^{2}}{4}+\frac{{(y+1)}^{2}}{16}=1;\) center: \((3,-1);\) vertices: \((3,-\text{5})\) and \((3,\text{3});\) co-vertices: \((1,-1)\) and \((5,-1);\) foci: \((3,-\text{1}-2\sqrt{3})\) and \((3,-\text{1+}2\sqrt{3})\)

  26. A large room in an art gallery is a whispering chamber. Its dimensions are 46 feet wide by 96 feet long as shown in .

    1. What is the standard form of the equation of the ellipse representing the outline of the room? Hint: assume a horizontal ellipse, and let the center of the room be the point \((0,0).\)
    2. If two visitors standing at the foci of this room can hear each other whisper, how far apart are the two visitors? Round to the nearest foot.
    Cevabı açıkla.
    1. We are assuming a horizontal ellipse with center \((0,0),\) so we need to find an equation of the form \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\) where \(a>b.\) We know that the length of the major axis, \(2a,\) is longer than the length of the minor axis, \(2b.\) So the length of the room, 96, is represented by the major axis, and the width of the room, 46, is represented by the minor axis.
      • Solving for \(a,\) we have \(2a=96,\) so \(a=48,\) and \({a}^{2}=2304.\)
      • Solving for \(b,\) we have \(2b=46,\) so \(b=23,\) and \({b}^{2}=529.\)

      Therefore, the equation of the ellipse is \(\frac{{x}^{2}}{2304}+\frac{{y}^{2}}{529}=1.\)

    2. To find the distance between the senators, we must find the distance between the foci, \((\pm c,0),\) where \({c}^{2}={a}^{2}-{b}^{2}.\) Solving for \(c,\) we have: \[\begin{array}{lllll}{c}^{2}={a}^{2}-{b}^{2} & \\ {c}^{2}=2304-529 & \begin{array}{llll} & & & \end{array}\text{Substitute using the values found in part (a)}. \\ c=\pm \sqrt{2304-529} & \begin{array}{llll} & & & \end{array}\text{Take the square root of both sides}. \\ c=\pm \sqrt{1775} & \begin{array}{llll} & & & \end{array}\text{Subtract}. \\ c\approx \pm 42 & \begin{array}{llll} & & & \end{array}\text{Round to the nearest foot}.\end{array}\]

      The points \((\pm 42,0)\) represent the foci. Thus, the distance between the senators is \(2(42)=84\) feet.

  27. Suppose a whispering chamber is 480 feet long and 320 feet wide.

    ⓐ What is the standard form of the equation of the ellipse representing the room? Hint: assume a horizontal ellipse, and let the center of the room be the point \((0,0).\)
    ⓑ If two people are standing at the foci of this room and can hear each other whisper, how far apart are the people? Round to the nearest foot.

    Cevabı açıkla.
    1. ⓐ \(\frac{{x}^{2}}{57,600}+\frac{{y}^{2}}{25,600}=1\)
    2. ⓑ The people are standing 358 feet apart.

  28. Define an ellipse in terms of its foci.

    Cevabı açıkla.

    An ellipse is the set of all points in the plane the sum of whose distances from two fixed points, called the foci, is a constant.

  29. Where must the foci of an ellipse lie?

  30. What special case of the ellipse do we have when the major and minor axis are of the same length?

    Cevabı açıkla.

    This special case would be a circle.

  31. For the special case mentioned in the previous question, what would be true about the foci of that ellipse?

  32. What can be said about the symmetry of the graph of an ellipse with center at the origin and foci along the y-axis?

    Cevabı açıkla.

    It is symmetric about the x-axis, y-axis, and the origin.

  33. \(2{x}^{2}+y=4\)

  34. \(4{x}^{2}+9{y}^{2}=36\)

    Cevabı açıkla.

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

  35. \(4{x}^{2}-{y}^{2}=4\)

  36. \(4{x}^{2}+9{y}^{2}=1\)

    Cevabı açıkla.

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

  37. \(4{x}^{2}-8x+9{y}^{2}-72y+112=0\)

  38. \(\frac{{x}^{2}}{4}+\frac{{y}^{2}}{49}=1\)

    Cevabı açıkla.

    \(\frac{{x}^{2}}{{2}^{2}}+\frac{{y}^{2}}{{7}^{2}}=1;\) Endpoints of major axis \((0,7)\) and \((0,-7).\) Endpoints of minor axis \((2,0)\) and \((-2,0).\) Foci at \((0,3\sqrt{5}),(0,-3\sqrt{5}).\)

  39. \(\frac{{x}^{2}}{100}+\frac{{y}^{2}}{64}=1\)

  40. \({x}^{2}+9{y}^{2}=1\)

    Cevabı açıkla.

    \(\frac{{x}^{2}}{{(1)}^{2}}+\frac{{y}^{2}}{{(\frac{1}{3})}^{2}}=1;\) Endpoints of major axis \((1,0)\) and \((-1,0).\) Endpoints of minor axis \((0,\frac{1}{3}),(0,-\frac{1}{3}).\) Foci at \((\frac{2\sqrt{2}}{3},0),(-\frac{2\sqrt{2}}{3},0).\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
|x|
absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
\binom{n}{k}
binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).
\pi
pi
Ratio of a circle's circumference to its diameter, 3.14159…
e
Euler's number
2.71828…, the base whose exponential is its own derivative.
i
imaginary unit
i² = −1.
\theta
theta
The usual name for an angle.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\mathbf{v},\ \vec{v}
vector
A quantity with magnitude and direction; a column of numbers.

How to: The Ellipse

  1. Write equations of ellipses in standard form.
  2. Graph ellipses centered at the origin.
  3. Graph ellipses not centered at the origin.
  4. Solve applied problems involving ellipses.
  5. Complete the square of a binomial expression. (IA 9.2.1)
  6. Graph a circle. (IA 11.1.4)
  7. Identify
  8. Find

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Kendini dene.

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Daha fazlası Precalculus

Complex numbersPolynomial functionsRational functionsSequences and seriesThe binomial theoremConic sectionsVectorsExponential and logarithmic functionsPolynomial division and the remainder theoremParametric equations and polar coordinates