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The Hyperbola

Locate a hyperbola’s vertices and foci.

The Hyperbola

  • Use the Distance Formula. (IA 11.1.1)
  • Graph a hyperbola with center at (0,0). (IA 11.4.1)
Example

Try it.

Use the distance formula to find the distance between the points (−5, −3) and (7,2).

Solution
Write the Distance Formula. \(d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\)
Label the points (–5, –3) as (x1, y1) and point (7, 2) as (x2, y2) and substitute. \(d=\sqrt{{\left(72-(-5)\right)}^{2}+{\left(2-(-3)\right)}^{2}}\)
Simplify. \(d=\sqrt{{12}^{2}+{5}^{2}}=\sqrt{144+55}=\sqrt{169}\)
\(d=13\)

Use the Distance Formula.

Try it.

Use the Distance Formula to find the distance between the points (−2,−5) and (−14,−10).

Try it.

Use the Distance Formula to find the distance between the points (10, −4) and (−1,5). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Condensed — the full section is in OpenStax College Algebra 2e.

Locating the Vertices and Foci of a Hyperbola

In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other. See .

Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points \((x,y)\) in a plane such that the difference of the distances between \((x,y)\) and the foci is a positive constant.

Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances.

As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle. See .

In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the x- and y-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin.

Condensed — the full section is in OpenStax College Algebra 2e.

Writing Equations of Hyperbolas in Standard Form

Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.

Condensed — the full section is in OpenStax College Algebra 2e.

Graphing Hyperbolas Centered at the Origin

When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form \(\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1\) for horizontal hyperbolas and the standard form \(\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1\) for vertical hyperbolas.

Condensed — the full section is in OpenStax College Algebra 2e.

Graphing Hyperbolas Not Centered at the Origin

Graphing hyperbolas centered at a point \((h,k)\) other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms \(\frac{{(x-h)}^{2}}{{a}^{2}}-\frac{{(y-k)}^{2}}{{b}^{2}}=1\) for horizontal hyperbolas, and \(\frac{{(y-k)}^{2}}{{a}^{2}}-\frac{{(x-h)}^{2}}{{b}^{2}}=1\) for vertical hyperbolas. From these standard form equations we can easily calculate and plot key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes.

Condensed — the full section is in OpenStax College Algebra 2e.

Solving Applied Problems Involving Hyperbolas

As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. See . For example, a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide!

The first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides.

Condensed — the full section is in OpenStax College Algebra 2e.

Key Equations

Hyperbola, center at origin, transverse axis on x-axis \(\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1\)
Hyperbola, center at origin, transverse axis on y-axis \(\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1\)
Hyperbola, center at \((h,k),\) transverse axis parallel to x-axis \(\frac{{(x-h)}^{2}}{{a}^{2}}-\frac{{(y-k)}^{2}}{{b}^{2}}=1\)
Hyperbola, center at \((h,k),\) transverse axis parallel to y-axis \(\frac{{(y-k)}^{2}}{{a}^{2}}-\frac{{(x-h)}^{2}}{{b}^{2}}=1\)

Key Concepts

  • A hyperbola is the set of all points \((x,y)\) in a plane such that the difference of the distances between \((x,y)\) and the foci is a positive constant.
  • The standard form of a hyperbola can be used to locate its vertices and foci. See .
  • When given the coordinates of the foci and vertices of a hyperbola, we can write the equation of the hyperbola in standard form. See and .
  • When given an equation for a hyperbola, we can identify its vertices, co-vertices, foci, asymptotes, and lengths and positions of the transverse and conjugate axes in order to graph the hyperbola. See and .
  • Real-world situations can be modeled using the standard equations of hyperbolas. For instance, given the dimensions of a natural draft cooling tower, we can find a hyperbolic equation that models its sides. 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 distance formula to find the distance between the points (−5, −3) and (7,2).

    Avslöja svaret
    Write the Distance Formula. \(d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\)
    Label the points (–5, –3) as (x1, y1) and point (7, 2) as (x2, y2) and substitute. \(d=\sqrt{{\left(72-(-5)\right)}^{2}+{\left(2-(-3)\right)}^{2}}\)
    Simplify. \(d=\sqrt{{12}^{2}+{5}^{2}}=\sqrt{144+55}=\sqrt{169}\)
    \(d=13\)

  2. Use the Distance Formula to find the distance between the points (−2,−5) and (−14,−10).

  3. Use the Distance Formula to find the distance between the points (10, −4) and (−1,5). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

  4. Graph \(\frac{{x}^{2}}{25}-\frac{{y}^{2}}{4}=1.\)

  5. Graph \(4{y}^{2}-16{x}^{2}=64.\)

    Avslöja svaret

    \(4{y}^{2}-16{x}^{2}=64\)
    To write the equation in standard form, divide
    each term by 64 to make the equation equal to 1.     		\(\frac{4{y}^{2}}{64}-\frac{16{x}^{2}}{64}=\frac{64}{64}\)
    Simplify. \(\ \frac{{y}^{2}}{16}-\frac{{x}^{2}}{4}=1\)
    Since the y 2 -term is positive, the transverse axis is vertical.
    Since \({a}^{2}=16\) then \(a=\text{\pm }4.\)
    The vertices are on the y -axis, \((0,\text{-}a),\) \((0,a).\)
    Since \({b}^{2}=4\) then \(b=\text{\pm }2.\)     		\((0,-4),\) \((0,4)\)
    Sketch the rectangle intersecting the x -axis at \((-2,0),\) \((2,0)\) and the y -axis at the vertices.
    Sketch the asymptotes through the diagonals of the rectangle.
    Draw the two branches of the hyperbola.

  6. Graph \(\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=1\) .

  7. Graph \(25{y}^{2}-9{x}^{2}=225\) .

  8. Identify the vertices and foci of the hyperbola with equation \(\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32}=1.\)

    Avslöja svaret

    The equation has the form \(\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1,\) so the transverse axis lies on the y-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set \(x=0,\) and solve for \(y.\)

    \[\begin{array}{l}1=\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32} \\ 1=\frac{{y}^{2}}{49}-\frac{{0}^{2}}{32} \\ 1=\frac{{y}^{2}}{49} \\ {y}^{2}=49 \\ y=\pm \sqrt{49}=\pm 7\end{array}\]

    The foci are located at \((0,\pm c).\) Solving for \(c,\)

    \[c=\sqrt{{a}^{2}+{b}^{2}}=\sqrt{49+32}=\sqrt{81}=9\]

    Therefore, the vertices are located at \((0,\pm 7),\) and the foci are located at \((0,\pm 9).\)

  9. Identify the vertices and foci of the hyperbola with equation \(\frac{{x}^{2}}{9}-\frac{{y}^{2}}{25}=1.\)

    Avslöja svaret

    Vertices: \((\pm 3,0);\) Foci: \((\pm \sqrt{34},0)\)

  10. What is the standard form equation of the hyperbola that has vertices \((\pm 6,0)\) and foci \((\pm 2\sqrt{10},0)?\)

    Avslöja svaret

    The vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the form \(\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1.\)

    The vertices are \((\pm 6,0),\) so \(a=6\) and \({a}^{2}=36.\)

    The foci are \((\pm 2\sqrt{10},0),\) so \(c=2\sqrt{10}\) and \({c}^{2}=40.\)

    Solving for \({b}^{2},\) we have

    \[\begin{array}{lllll}{b}^{2}={c}^{2}-{a}^{2} & \\ {b}^{2}=40-36 & \begin{array}{llll} & & & \end{array}\text{Substitute for }{c}^{2}\ \text{and }{a}^{2}. \\ {b}^{2}=4 & \begin{array}{llll} & & & \end{array}\text{Subtract}.\end{array}\]

    Finally, we substitute \({a}^{2}=36\) and \({b}^{2}=4\) into the standard form of the equation, \(\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1.\) The equation of the hyperbola is \(\frac{{x}^{2}}{36}-\frac{{y}^{2}}{4}=1,\) as shown in .

  11. What is the standard form equation of the hyperbola that has vertices \((0,\pm 2)\) and foci \((0,\pm 2\sqrt{5})?\)

    Avslöja svaret

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

  12. What is the standard form equation of the hyperbola that has vertices at \((0,-2)\) and \((6,-2)\) and foci at \((-2,-2)\) and \((8,-2)?\)

    Avslöja svaret

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

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

    First, we identify the center, \((h,k).\) The center is halfway between the vertices \((0,-2)\) and \((6,-2).\) Applying the midpoint formula, we have

    \[(h,k)=(\frac{0+6}{2},\frac{-2+(-2)}{2})=(3,-2)\]

    Next, we find \({a}^{2}.\) The length of the transverse axis, \(2a,\) is bounded by the vertices. So, we can find \({a}^{2}\) by finding the distance between the x-coordinates of the vertices.

    \[\begin{array}{l}2a=|0-6| \\ 2a=6 \\ \ a=3 \\ {a}^{2}=9\end{array}\]

    Now we need to find \({c}^{2}.\) The coordinates of the foci are \((h\pm c,k).\) So \((h-c,k)=(-2,-2)\) and \((h+c,k)=(8,-2).\) We can use the x-coordinate from either of these points to solve for \(c.\) Using the point \((8,-2),\) and substituting \(h=3,\)

    \[\begin{array}{l}h+c=8 \\ 3+c=8 \\ \ c=5 \\ \ {c}^{2}=25\end{array}\]

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

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

    Finally, substitute the values found for \(h,k,{a}^{2},\) and \({b}^{2}\) into the standard form of the equation.

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

  13. What is the standard form equation of the hyperbola that has vertices \((1,-2)\) and \((1,\text{8})\) and foci \((1,-10)\) and \((1,16)?\)

    Avslöja svaret

    \(\frac{{(y-3)}^{2}}{25}-\frac{{(x-1)}^{2}}{144}=1\)

  14. Graph the hyperbola given by the equation \(\frac{{y}^{2}}{64}-\frac{{x}^{2}}{36}=1.\) Identify and label the vertices, co-vertices, foci, and asymptotes.

    Avslöja svaret

    The standard form that applies to the given equation is \(\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1.\) Thus, the transverse axis is on the y-axis

    The coordinates of the vertices are \((0,\pm a)=(0,\pm \sqrt{64})=(0,\pm 8)\)

    The coordinates of the co-vertices are \((\pm b,0)=(\pm \sqrt{36},\ 0)=(\pm 6,0)\)

    The coordinates of the foci are \((0,\pm c),\) where \(c=\pm \sqrt{{a}^{2}+{b}^{2}}.\) Solving for \(c,\) we have

    \[c=\pm \sqrt{{a}^{2}+{b}^{2}}=\pm \sqrt{64+36}=\pm \sqrt{100}=\pm 10\]

    Therefore, the coordinates of the foci are \((0,\pm 10)\)

    The equations of the asymptotes are \(y=\pm \frac{a}{b}x=\pm \frac{8}{6}x=\pm \frac{4}{3}x\)

    Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola, as shown in .

  15. Graph the hyperbola given by the equation \(\frac{{x}^{2}}{144}-\frac{{y}^{2}}{81}=1.\) Identify and label the vertices, co-vertices, foci, and asymptotes.

    Avslöja svaret

    vertices: \((\pm 12,0);\) co-vertices: \((0,\pm 9);\) foci: \((\pm 15,0);\) asymptotes: \(y=\pm \frac{3}{4}x;\)

  16. Graph the hyperbola given by the equation \(9{x}^{2}-4{y}^{2}-36x-40y-388=0.\) Identify and label the center, vertices, co-vertices, foci, and asymptotes.

    Avslöja svaret

    Start by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation.

    \[(9{x}^{2}-36x)-(4{y}^{2}+40y)=388\]

    Factor the leading coefficient of each expression.

    \[9({x}^{2}-4x)-4({y}^{2}+10y)=388\]

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

    \[9({x}^{2}-4x+4)-4({y}^{2}+10y+25)=388+36-100\]

    Rewrite as perfect squares.

    \[9{(x-2)}^{2}-4{(y+5)}^{2}=324\]

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

    \[\frac{{(x-2)}^{2}}{36}-\frac{{(y+5)}^{2}}{81}=1\]

    The standard form that applies to the given equation is \(\frac{{(x-h)}^{2}}{{a}^{2}}-\frac{{(y-k)}^{2}}{{b}^{2}}=1,\) where \({a}^{2}=36\) and \({b}^{2}=81,\) or \(a=6\) and \(b=9.\) Thus, the transverse axis is parallel to the x-axis. It follows that:

    • the center of the ellipse is \((h,k)=(2,-5)\)
    • the coordinates of the vertices are \((h\pm a,k)=(2\pm 6,-5),\) or \((-4,-5)\) and \((8,-5)\)
    • the coordinates of the co-vertices are \((h,k\pm b)=(2,-5\pm 9),\) or \((2,-14)\) and \((2,4)\)
    • the coordinates of the foci are \((h\pm c,k),\) where \(c=\pm \sqrt{{a}^{2}+{b}^{2}}.\) Solving for \(c,\) we have

    \[c=\pm \sqrt{36+81}=\pm \sqrt{117}=\pm 3\sqrt{13}\]

    Therefore, the coordinates of the foci are \((2-3\sqrt{13},-5)\) and \((2+3\sqrt{13},-5).\)

    The equations of the asymptotes are \(y=\pm \frac{b}{a}(x-h)+k=\pm \frac{3}{2}(x-2)-5.\)

    Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola, as shown in .

  17. Graph the hyperbola given by the standard form of an equation \(\frac{{(y+4)}^{2}}{100}-\frac{{(x-3)}^{2}}{64}=1.\) Identify and label the center, vertices, co-vertices, foci, and asymptotes.

    Avslöja svaret

    center: \((3,-4);\) vertices: \((3,-14)\) and \((3,6);\) co-vertices: \((-5,-4);\) and \((11,-4);\) foci: \((3,-4-2\sqrt{41})\) and \((3,-4+2\sqrt{41});\) asymptotes: \(y=\pm \frac{5}{4}(x-3)-4\)

  18. The design layout of a cooling tower is shown in . The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart.

    Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

    Avslöja svaret

    We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hyperbola centered at the origin: \(\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1,\) where the branches of the hyperbola form the sides of the cooling tower. We must find the values of \({a}^{2}\) and \({b}^{2}\) to complete the model.

    First, we find \({a}^{2}.\) Recall that the length of the transverse axis of a hyperbola is \(2a.\) This length is represented by the distance where the sides are closest, which is given as \(\ 60\) meters. So, \(2a=60.\) Therefore, \(a=30\) and \({a}^{2}=900.\)

    To solve for \({b}^{2},\) we need to substitute for \(x\) and \(y\) in our equation using a known point. To do this, we can use the dimensions of the tower to find some point \((x,y)\) that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the y-axis bisects the tower, our x-value can be represented by the radius of the top, or 36 meters. The y-value is represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore,

    \[\begin{array}{lllll}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1 & \begin{array}{llll} & & & \end{array}\text{Standard form of horizontal hyperbola}. \\ \ {b}^{2}=\frac{{y}^{2}}{\frac{{x}^{2}}{{a}^{2}}-1} & \begin{array}{llll} & & & \end{array}\text{Isolate }{b}^{2} \\ \ =\frac{{(79.6)}^{2}}{\frac{{(36)}^{2}}{900}-1} & \begin{array}{llll} & & & \end{array}\text{Substitute for }{a}^{2},x,\ \text{and }y \\ \ \approx 14400.3636 & \begin{array}{llll} & & & \end{array}\text{Round to four decimal places}\end{array}\]

    The sides of the tower can be modeled by the hyperbolic equation

    \[\frac{{x}^{2}}{900}-\frac{{y}^{2}}{14400.3636}=1,\text{or}\ \frac{{x}^{2}}{{30}^{2}}-\frac{{y}^{2}}{{120.0015}^{2}}=1\]

  19. A design for a cooling tower project is shown in . Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

    Avslöja svaret

    The sides of the tower can be modeled by the hyperbolic equation. \(\frac{{x}^{2}}{400}-\frac{{y}^{2}}{3600}=1\text{or }\frac{{x}^{2}}{{20}^{2}}-\frac{{y}^{2}}{{60}^{2}}=1.\)

  20. Define a hyperbola in terms of its foci.

    Avslöja svaret

    A hyperbola is the set of points in a plane the difference of whose distances from two fixed points (foci) is a positive constant.

  21. What can we conclude about a hyperbola if its asymptotes intersect at the origin?

  22. What must be true of the foci of a hyperbola?

    Avslöja svaret

    The foci must lie on the transverse axis and be in the interior of the hyperbola.

  23. If the transverse axis of a hyperbola is vertical, what do we know about the graph?

  24. Where must the center of hyperbola be relative to its foci?

    Avslöja svaret

    The center must be the midpoint of the line segment joining the foci.

  25. \(3{y}^{2}+2x=6\)

  26. \(\frac{{x}^{2}}{36}-\frac{{y}^{2}}{9}=1\)

    Avslöja svaret

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

  27. \(5{y}^{2}+4{x}^{2}=6x\)

  28. \(25{x}^{2}-16{y}^{2}=400\)

    Avslöja svaret

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

  29. \(-9{x}^{2}+18x+{y}^{2}+4y-14=0\)

  30. \(\frac{{x}^{2}}{25}-\frac{{y}^{2}}{36}=1\)

    Avslöja svaret

    \(\frac{{x}^{2}}{{5}^{2}}-\frac{{y}^{2}}{{6}^{2}}=1;\) vertices: \((5,0),(-5,0);\) foci: \((\sqrt{61},0),(-\sqrt{61},0);\) asymptotes: \(y=\frac{6}{5}x,y=-\frac{6}{5}x\)

  31. \(\frac{{x}^{2}}{100}-\frac{{y}^{2}}{9}=1\)

  32. \(\frac{{y}^{2}}{4}-\frac{{x}^{2}}{81}=1\)

    Avslöja svaret

    \(\frac{{y}^{2}}{{2}^{2}}-\frac{{x}^{2}}{{9}^{2}}=1;\) vertices: \((0,2),(0,-2);\) foci: \((0,\sqrt{85}),(0,-\sqrt{85});\) asymptotes: \(y=\frac{2}{9}x,y=-\frac{2}{9}x\)

  33. \(9{y}^{2}-4{x}^{2}=1\)

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

    Avslöja svaret

    \(\frac{{(x-1)}^{2}}{{3}^{2}}-\frac{{(y-2)}^{2}}{{4}^{2}}=1;\) vertices: \((4,2),(-2,2);\) foci: \((6,2),(-4,2);\) asymptotes: \(y=\frac{4}{3}(x-1)+2,y=-\frac{4}{3}(x-1)+2\)

  35. \(\frac{{(y-6)}^{2}}{36}-\frac{{(x+1)}^{2}}{16}=1\)

  36. \(\frac{{(x-2)}^{2}}{49}-\frac{{(y+7)}^{2}}{49}=1\)

    Avslöja svaret

    \(\frac{{(x-2)}^{2}}{{7}^{2}}-\frac{{(y+7)}^{2}}{{7}^{2}}=1;\) vertices: \((9,-7),(-5,-7);\) foci: \((2+7\sqrt{2},-7),(2-7\sqrt{2},-7);\) asymptotes: \(y=x-9,y=-x-5\)

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

  38. \(-9{x}^{2}-54x+9{y}^{2}-54y+81=0\)

    Avslöja svaret

    \(\frac{{(x+3)}^{2}}{{3}^{2}}-\frac{{(y-3)}^{2}}{{3}^{2}}=1;\) vertices: \((0,3),(-6,3);\) foci: \((-3+3\sqrt{2},1),(-3-3\sqrt{2},1);\) asymptotes: \(y=x+6,y=-x\)

  39. \(4{x}^{2}-24x-36{y}^{2}-360y+864=0\)

  40. \(-4{x}^{2}+24x+16{y}^{2}-128y+156=0\)

    Avslöja svaret

    \(\frac{{(y-4)}^{2}}{{2}^{2}}-\frac{{(x-3)}^{2}}{{4}^{2}}=1;\) vertices: \((3,6),(3,2);\) foci: \((3,4+2\sqrt{5}),(3,4-2\sqrt{5});\) asymptotes: \(y=\frac{1}{2}(x-3)+4,y=-\frac{1}{2}(x-3)+4\)

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.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\neq
not equal
The two sides are different.
|x|
absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
i
imaginary unit
i² = −1.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}
number sets
Naturals, integers, rationals, reals, complex numbers.

How to: The Hyperbola

  1. Locate a hyperbola’s vertices and foci.
  2. Write equations of hyperbolas in standard form.
  3. Graph hyperbolas centered at the origin.
  4. Graph hyperbolas not centered at the origin.
  5. Solve applied problems involving hyperbolas.
  6. Use the Distance Formula. (IA 11.1.1)
  7. Graph a hyperbola with center at (0,0). (IA 11.4.1)
  8. Write the equation in standard form.

Questions people ask

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To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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

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