Foci calculator hyperbola

Foci calculator hyperbola DEFAULT

Learning Outcomes

  • Derive an equation for a hyperbola centered at the origin
  • Write an equation for a hyperbola centered at the origin
  • Solve an applied problem involving hyperbolas

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.

A hyperbola

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 [latex]\left(x,y\right)[/latex] in a plane such that the difference of the distances between [latex]\left(x,y\right)[/latex] 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.

Key features of the hyperbola

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.

Standard Form of the Equation of a Hyperbola Centered at the Origin

Let [latex]\left(-c,0\right)[/latex] and [latex]\left(c,0\right)[/latex] be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points [latex]\left(x,y\right)[/latex] such that the difference of the distances from [latex]\left(x,y\right)[/latex] to the foci is constant.

If [latex]\left(a,0\right)[/latex] is a vertex of the hyperbola, the distance from [latex]\left(-c,0\right)[/latex] to [latex]\left(a,0\right)[/latex] is [latex]a-\left(-c\right)=a+c[/latex]. The distance from [latex]\left(c,0\right)[/latex] to [latex]\left(a,0\right)[/latex] is [latex]c-a[/latex]. The difference of the distances from the foci to the vertex is

[latex]\left(a+c\right)-\left(c-a\right)=2a[/latex]

If [latex]\left(x,y\right)[/latex] is a point on the hyperbola, we can define the following variables:

[latex]\begin{align}&{d}_{2}=\text{the distance from }\left(-c,0\right)\text{ to }\left(x,y\right)\\ &{d}_{1}=\text{the distance from }\left(c,0\right)\text{ to }\left(x,y\right)\end{align}[/latex]

By definition of a hyperbola, [latex]\lvert{d}_{2}-{d}_{1}\rvert[/latex] is constant for any point [latex]\left(x,y\right)[/latex] on the hyperbola. We know that the difference of these distances is [latex]2a[/latex] for the vertex [latex]\left(a,0\right)[/latex]. It follows that [latex]\lvert{d}_{2}-{d}_{1}\rvert=2a[/latex] for any point on the hyperbola. The derivation of the equation of a hyperbola is based on applying the distance formula, but is again beyond the scope of this text. The standard form of an equation of a hyperbola centered at the origin with vertices [latex]\left(\pm a,0\right)[/latex] and co-vertices [latex]\left(0\pm b\right)[/latex] is [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex].

A General Note: Standard Forms of the Equation of a Hyperbola with Center (0,0)

The standard form of the equation of a hyperbola with center [latex]\left(0,0\right)[/latex] and transverse axis on the x-axis is

[latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex]

where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(\pm a,0\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(0,\pm b\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(\pm c,0\right)[/latex]
  • the equations of the asymptotes are [latex]y=\pm \dfrac{b}{a}x[/latex]

The standard form of the equation of a hyperbola with center [latex]\left(0,0\right)[/latex] and transverse axis on the y-axis is

[latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex]

where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(0,\pm a\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(\pm b,0\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(0,\pm c\right)[/latex]
  • the equations of the asymptotes are [latex]y=\pm \dfrac{a}{b}x[/latex]

Note that the vertices, co-vertices, and foci are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

(a) Horizontal hyperbola with center [latex]\left(0,0\right)[/latex] (b) Vertical hyperbola with center [latex]\left(0,0\right)[/latex]

How To: Given the equation of a hyperbola in standard form, locate its vertices and foci.

  1. Determine whether the transverse axis lies on the x– or y-axis. Notice that [latex]{a}^{2}[/latex] is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.

a. If the equation has the form [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex], then the transverse axis lies on the x-axis. The vertices are located at [latex]\left(\pm a,0\right)[/latex], and the foci are located at [latex]\left(\pm c,0\right)[/latex].

b. If the equation has the form [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex], then the transverse axis lies on the y-axis. The vertices are located at [latex]\left(0,\pm a\right)[/latex], and the foci are located at [latex]\left(0,\pm c\right)[/latex].

  1. Solve for [latex]a[/latex] using the equation [latex]a=\sqrt{{a}^{2}}[/latex].
  2. Solve for [latex]c[/latex] using the equation [latex]c=\sqrt{{a}^{2}+{b}^{2}}[/latex].

Example: Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the hyperbola with equation [latex]\dfrac{{y}^{2}}{49}-\dfrac{{x}^{2}}{32}=1[/latex].

Show Solution

The equation has the form [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex], 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 [latex]x=0[/latex], and solve for [latex]y[/latex].

[latex]\begin{align}1=\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32} \\[1mm] 1=\frac{{y}^{2}}{49}-\frac{{0}^{2}}{32} \\[1mm] 1=\frac{{y}^{2}}{49} \\[1mm] {y}^{2}=49 \\[1mm] y=\pm \sqrt{49}=\pm 7 \end{align}[/latex]

The foci are located at [latex]\left(0,\pm c\right)[/latex]. Solving for [latex]c[/latex],

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

Therefore, the vertices are located at [latex]\left(0,\pm 7\right)[/latex], and the foci are located at [latex]\left(0,9\right)[/latex].

Try It

Identify the vertices and foci of the hyperbola with equation [latex]\dfrac{{x}^{2}}{9}-\dfrac{{y}^{2}}{25}=1[/latex].

Show Solution

Vertices: [latex]\left(\pm 3,0\right)[/latex]; Foci: [latex]\left(\pm \sqrt{34},0\right)[/latex]

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.

Hyperbolas Centered at the Origin

Reviewing the standard forms given for hyperbolas centered at [latex]\left(0,0\right)[/latex], we see that the vertices, co-vertices, and foci are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. Note that this equation can also be rewritten as [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex]. This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.

How To: Given the vertices and foci of a hyperbola centered at [latex]\left(0,\text{0}\right)[/latex], write its equation in standard form.

  1. Determine whether the transverse axis lies on the x– or y-axis.
    1. If the given coordinates of the vertices and foci have the form [latex]\left(\pm a,0\right)[/latex] and [latex]\left(\pm c,0\right)[/latex], respectively, then the transverse axis is the x-axis. Use the standard form [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex].
    2. If the given coordinates of the vertices and foci have the form [latex]\left(0,\pm a\right)[/latex] and [latex]\left(0,\pm c\right)[/latex], respectively, then the transverse axis is the y-axis. Use the standard form [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex].
  2. Find [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex].
  3. Substitute the values for [latex]{a}^{2}[/latex] and [latex]{b}^{2}[/latex] into the standard form of the equation determined in Step 1.

Example: Finding the Equation of a Hyperbola Centered at (0,0) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices [latex]\left(\pm 6,0\right)[/latex] and foci [latex]\left(\pm 2\sqrt{10},0\right)?[/latex]

Show Solution

The vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the form [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex].

The vertices are [latex]\left(\pm 6,0\right)[/latex], so [latex]a=6[/latex] and [latex]{a}^{2}=36[/latex].

The foci are [latex]\left(\pm 2\sqrt{10},0\right)[/latex], so [latex]c=2\sqrt{10}[/latex] and [latex]{c}^{2}=40[/latex].

Solving for [latex]{b}^{2}[/latex], we have

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

Finally, we substitute [latex]{a}^{2}=36[/latex] and [latex]{b}^{2}=4[/latex] into the standard form of the equation, [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex]. The equation of the hyperbola is [latex]\dfrac{{x}^{2}}{36}-\dfrac{{y}^{2}}{4}=1[/latex].

Try It

What is the standard form equation of the hyperbola that has vertices [latex]\left(0,\pm 2\right)[/latex] and foci [latex]\left(0,\pm 2\sqrt{5}\right)?[/latex]

Show Solution

[latex]\dfrac{{y}^{2}}{4}-\dfrac{{x}^{2}}{16}=1[/latex]

Hyperbolas Not Centered at the Origin

Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated [latex]h[/latex] units horizontally and [latex]k[/latex] units vertically, the center of the hyperbola will be [latex]\left(h,k\right)[/latex]. This translation results in the standard form of the equation we saw previously, with [latex]x[/latex] replaced by [latex]\left(x-h\right)[/latex] and [latex]y[/latex] replaced by [latex]\left(y-k\right)[/latex].

A General Note: Standard Forms of the Equation of a Hyperbola with Center (h, k)

The standard form of the equation of a hyperbola with center [latex]\left(h,k\right)[/latex] and transverse axis parallel to the x-axis is
[latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex]
where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(h\pm a,k\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex]

The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is [latex]2a[/latex] and its width is [latex]2b[/latex]. The slopes of the diagonals are [latex]\pm \frac{b}{a}[/latex], and each diagonal passes through the center [latex]\left(h,k\right)[/latex]. Using the point-slope formula, it is simple to show that the equations of the asymptotes are [latex]y=\pm \frac{b}{a}\left(x-h\right)+k[/latex].

The standard form of the equation of a hyperbola with center [latex]\left(h,k\right)[/latex] and transverse axis parallel to the y-axis is
[latex]\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}-\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex]
where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(h,k\pm a\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(h\pm b,k\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(h,k\pm c\right)[/latex]

Using the reasoning above, the equations of the asymptotes are [latex]y=\pm \frac{a}{b}\left(x-h\right)+k[/latex].

(a) Horizontal hyperbola with center [latex]\left(h,k\right)[/latex] (b) Vertical hyperbola with center [latex]\left(h,k\right)[/latex]

Like hyperbolas centered at the origin, hyperbolas centered at a point [latex]\left(h,k\right)[/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.

How To: Given the vertices and foci of a hyperbola centered at [latex]\left(h,k\right)[/latex], write its equation in standard form.

  1. Determine whether the transverse axis is parallel to the x– or y-axis.
    1. If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form [latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex].
    2. If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form [latex]\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}-\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex].
  2. Identify the center of the hyperbola, [latex]\left(h,k\right)[/latex], using the midpoint formula and the given coordinates for the vertices.
  3. Find [latex]{a}^{2}[/latex] by solving for the length of the transverse axis, [latex]2a[/latex] , which is the distance between the given vertices.
  4. Find [latex]{c}^{2}[/latex] using [latex]h[/latex] and [latex]k[/latex] found in Step 2 along with the given coordinates for the foci.
  5. Solve for [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex].
  6. Substitute the values for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form of the equation determined in Step 1.

Example: Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices at [latex]\left(0,-2\right)[/latex] and [latex]\left(6,-2\right)[/latex] and foci at [latex]\left(-2,-2\right)[/latex] and [latex]\left(8,-2\right)?[/latex]

Show Solution

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

 

[latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex]

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

[latex]\left(h,k\right)=\left(\dfrac{0+6}{2},\dfrac{-2+\left(-2\right)}{2}\right)=\left(3,-2\right)[/latex]

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

[latex]\begin{gathered}2a=|0 - 6| \\ 2a=6 \\ a=3 \\ {a}^{2}=9 \end{gathered}[/latex]

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

[latex]\begin{gathered}h+c=8 \\ 3+c=8 \\ c=5 \\ {c}^{2}=25 \end{gathered}[/latex]

Next, solve for [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}:[/latex]

[latex]\begin{align}{b}^{2}&={c}^{2}-{a}^{2} \\ &=25 - 9 \\ &=16 \end{align}[/latex]

Finally, substitute the values found for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form of the equation.

[latex]\dfrac{{\left(x - 3\right)}^{2}}{9}-\dfrac{{\left(y+2\right)}^{2}}{16}=1[/latex]

Try It

What is the standard form equation of the hyperbola that has vertices [latex]\left(1,-2\right)[/latex] and [latex]\left(1,\text{8}\right)[/latex] and foci [latex]\left(1,-10\right)[/latex] and [latex]\left(1,16\right)?[/latex]

Show Solution

[latex]\dfrac{{\left(y - 3\right)}^{2}}{25}-\dfrac{{\left(x - 1\right)}^{2}}{144}=1[/latex]

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. For example a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide!

description in caption

Cooling towers at the Drax power station in North Yorkshire, United Kingdom (credit: Les Haines, Flickr)

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 Example 6 we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides.

Example: Solving Applied Problems Involving Hyperbolas

The design layout of a cooling tower is shown below. 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.

Project design for a natural draft cooling tower

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.

Show Solution

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: [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex], where the branches of the hyperbola form the sides of the cooling tower. We must find the values of [latex]{a}^{2}[/latex] and [latex]{b}^{2}[/latex] to complete the model.

First, we find [latex]{a}^{2}[/latex]. Recall that the length of the transverse axis of a hyperbola is [latex]2a[/latex]. This length is represented by the distance where the sides are closest, which is given as [latex]\text{ }65.3\text{ }[/latex] meters. So, [latex]2a=60[/latex]. Therefore, [latex]a=30[/latex] and [latex]{a}^{2}=900[/latex].

To solve for [latex]{b}^{2}[/latex], we need to substitute for [latex]x[/latex] and [latex]y[/latex] in our equation using a known point. To do this, we can use the dimensions of the tower to find some point [latex]\left(x,y\right)[/latex] 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,

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

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

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

Try It

A design for a cooling tower project is shown below. 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.

Show Solution

The sides of the tower can be modeled by the hyperbolic equation. [latex]\dfrac{{x}^{2}}{400}-\dfrac{{y}^{2}}{3600}=1\text{ or }\dfrac{{x}^{2}}{{20}^{2}}-\dfrac{{y}^{2}}{{60}^{2}}=1[/latex].

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Conics: Hyperbolas: Finding Information
     From the Equation
(page 2 of 3)

Sections: Introduction, Finding information from the equation, Finding the equation from information


  • Find the center, vertices, foci, eccentricity, and asymptotes of the hyperbola with the given equation, and sketch:  (y^2) / 25 - (x^2) / 144 = 1
  • Since the y part of the equation is added, then the center, foci, and vertices will be above and below the center (on a line paralleling the y-axis), rather than side by side.

    Looking at the denominators, I see that a2 = 25 and b2 = 144, so a = 5 and b = 12. The equation c2 – a2 = b2 tells me that c2 = 144 + 25 = 169, so c = 13, and the eccentricity is e = 13/5. Since x2 = (x – 0)2 and y2 = (y – 0)2, then the center is at (h, k) = (0, 0). The vertices and foci are above and below the center, so the foci are at (0, –13) and (0, 13), and the vertices are at (0, 5) and (0, –5).

     
    Because the y part of the equation is dominant (being added, not subtracted), then the slope of the asymptotes has the a on top, so the slopes will be m = ± 5/12. To graph, I start with the center, and draw the asymptotes through it, using dashed lines:

     

     

    axis system, from -50 to 50, with center and asymptotes marked 

     

     
    Then I draw in the vertices, and rough in the graph, rotating the paper as necessary and "eye-balling" for smoothness:

     

    vertices added, and branches roughed in

     

     

    Then I draw in the final graph as a neat, smooth, heavier line:

     

    final graph, with 'answer' line drawn bold and clear, but neatly

    And the rest of my answer is:

      center (0, 0), vertices (0, –5) and (0, 5), foci (0, –13) and (0, 13),
      eccentricity e = 13/5, and asymptotes y = +/- (5/12) x

  • Give the center, vertices, foci, and asymptotes for the hyperbola
    with equation:   [(x + 3)^2] / 16 - [(y - 2)^2] / 9 = 1
  • Since the x part is added, then a2 = 16 and b2 = 9, so a = 4 and b = 3. Also, this hyperbola's foci and vertices are to the left and right of the center, on a horizontal line paralleling the x-axis.

    From the equation, clearly the center is at (h, k) = (–3, 2). Since the vertices are a = 4 units to either side, then they are at (–7, 2) and at (1, 2). The equation c2 – a2 = b2 gives me c2 = 9 + 16 = 25, so c = 5, and the foci, being 5 units to either side of the center, must be at (–8, 2) and (2, 2). Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved

    Since the a2 went with the x part of the equation, then a is in the denominator of the slopes of the asymptotes, giving me m = ± 3/4. Keeping in mind that the asymptotes go through the center of the hyperbola, the asymptes are then given by the straight-line equations  y – 2 = ± (3/4)(x + 3).

      center (–3, 2), vertices (–7, 2) and (1, 2), foci (–8, 2) and (2, 2),
      and asymptotes y = +/- (3/4) (x + 3) + 2

  • Find the center, vertices, and asymptotes of the hyperbola with equation
    4x2 – 5y2 + 40x – 30y – 45 = 0.
  • To find the information I need, I'll first have to convert this equation to "conics" form by completing the square.

      4x2 + 40x – 5y2 – 30y = 45
      4(x2 + 10x        ) – 5(y2 + 6y       ) = 45 + 4(   ) – 5(  )
      4(x2 + 10x + 25) – 5(y2 + 6y + 9) = 45 + 4(25) – 5(9)
      4(x + 5)2 – 5(y + 3)2 = 45 + 100 – 45

      [4(x + 5)^2] / 100 - [5(y + 3)^2] / 100 = 100 / 100

      [(x + 5)^2] / 25 - [(y +3)^2] / 20 = 1

    Then the center is at (h, k) = (–5, –3). Since the x part of the equation is added, then the center, foci, and vertices lie on a horizontal line paralleling the x-axis; a2 = 25 and b2 = 20, so a = 5 and b = 2sqrt[5]. The equation a2 + b2 = c2 gives me c2 = 25 + 20 = 45, so c = sqrt[45] = 3sqrt[5]. The slopes of the two asymptotes will be m = ± (2/5)sqrt[5]. Then my complete answer is:

      center (–5, –3), vertices (–10, –3) and (0, –3),

      foci (-5 - 3sqrt[5], -3) and (-5 + 3sqrt[5], -3),

      and asymptotes y = +/- (2sqrt[5] / 5) (x + 5) - 3

If I had needed to graph this hyperbola, I'd have used a decimal approximation of ± 0.89442719... for the slope, but would have rounded the value to something reasonable like m = ± 0.9.

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Cite this article as:

Stapel, Elizabeth. "Conics: Hyperbolas: Finding Information From the Equation." Purplemath.
    Available from https://www.purplemath.com/modules/hyperbola2.htm.
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Video transcript

Sours: https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:conics/x9e81a4f98389efdf:hyperb-foci/v/foci-of-a-hyperbola

Hyperbola foci calculator

Hyperbola Calculator


Given the hyperbola below, calculate the equation of the asymptotes, intercepts, foci points, eccentricity and other items.

Since our first variable is y, the hyperbola has a verticaltransverse axis or North-Southopening

Determine the equation of the asymptotes:

a = √100
a = 10

b = √49
b = 7

Asymptote 1  =  ax
  b

Asymptote 1  =  10x
  7

Asymptote 2  =  -ax
  b

Asymptote 2  =  -10x
  7

Determine y-intercepts:

y-intercepts = ±a
y-intercepts = ±10
y-intercepts =(0, 10) and (0, -10)

Determine the foci:

Our foci are at (0,c) and (0,-c) where
a2+ b2= c2
Therefore, c = √a2 + b2

a = √102 + 72
c = √100 + 49
c = √149
c = 12.206555615734
Therefore, our foci are at (0,12.206555615734) and (0,-12.206555615734)

Calculate eccentricity ε

ε  =  c
  a

ε  =  12.206555615734
  10

ε = 1.2206555615734

Calculate latus rectum:

Latus Rectum  =  2b2
  a

Latus Rectum  =  2(7)2
  10

Latus Rectum  =  2(49)
  10

Latus Rectum  =  98
  10

Latus Rectum = 9.8

Calculate semi-latus rectum l:

l  =  Latus Rectum
  2

l  =  9.8
  2

l = 4.9


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Writing the equation of a hyperbola given the foci and vertices

Hyperbola Calculator

Answer

Equation of the hyperbola: $$$x^{2} - 4 y^{2}=49$$$ or $$$x^{2} - 4 y^{2} - 49=0$$$.

Graph: to graph the hyperbola, visit the hyperbola graphing calculator (choose the "Implicit" option).

Standard form: $$$\frac{x^{2}}{49} - \frac{4 y^{2}}{49}=1$$$.

Center: $$$\left(0,0\right)$$$.

Vertices: $$$\left(-7,0\right)$$$, $$$\left(7,0\right)$$$.

Co-vertices: $$$\left(0,- \frac{7}{2}\right)$$$, $$$\left(0,\frac{7}{2}\right)$$$.

Foci: $$$\left(- \frac{7 \sqrt{5}}{2},0\right)\approx \left(-7.82623792124926,0\right)$$$, $$$\left(\frac{7 \sqrt{5}}{2},0\right)\approx \left(7.82623792124926,0\right)$$$.

Eccentricity: $$$\frac{\sqrt{5}}{2}\approx 1.11803398874989$$$.

Linear eccentricity: $$$\frac{7 \sqrt{5}}{2}\approx 7.82623792124926$$$.

Focal Parameter: $$$\frac{7 \sqrt{5}}{10}\approx 1.56524758424985$$$.

Major (transverse) axis length: $$$14$$$.

Semimajor axis length: $$$7$$$.

Minor (conjugate) axis length: $$$7$$$.

Semiminor axis length: $$$\frac{7}{2}$$$.

First asymptote: $$$y=- \frac{x}{2}$$$

Second asymptote: $$$y=\frac{x}{2}$$$

First directrix: $$$x=- \frac{14 \sqrt{5}}{5}\approx -6.26099033699941$$$.

Second directrix: $$$x=\frac{14 \sqrt{5}}{5}\approx 6.26099033699941$$$.

First latus rectum: $$$x=- \frac{7 \sqrt{5}}{2}\approx -7.82623792124926$$$.

Second latus rectum: $$$x=\frac{7 \sqrt{5}}{2}\approx 7.82623792124926$$$.

The length of the latera recta: $$$\frac{7}{2}$$$.

x-intercept: $$$\left(0, 0\right)$$$.

y-intercept: $$$\left(0, 0\right)$$$.

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