A hyperbola is a type of conic section made up of two curves that resemble parabolas (although they are not). These pairs of curves, also called branches, can either open up and down or left and right. Additionally, each curve contains a vertex.
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Jetzt kostenlos anmeldenA hyperbola is a type of conic section made up of two curves that resemble parabolas (although they are not). These pairs of curves, also called branches, can either open up and down or left and right. Additionally, each curve contains a vertex.
In this article, we aim to look at the properties of hyperbolas and identity equations that describe such conic sections. By familiarizing ourselves with these concepts, we can ultimately graph hyperbolas given a pair of coordinates.
A hyperbola is the set H of all points P in a plane where the absolute value of the difference in distance between two fixed points, called the foci, F1 and F2, is constant, k, that is
.
We shall further look into the components of this graph in the following section.
To visualize this, observe the graph below.
Notice that unlike other types of conic sections, hyperbolas are made up of two branches rather than one.
From a geometric point of view, a hyperbola is produced when a plane cuts parallel to the cone axis of a double-napped cone.
Let's explore more closely a graphical representation of a hyperbola. Here, we introduce more vital components that make up a hyperbola.
Every hyperbola has two axes of symmetry: the transverse axis and the conjugate axis
The conjugate axis (the y-axis) is a line perpendicular to the transverse axis and contain the co-vertices.
The transverse axis (the x-axis) is a line that passes through the centre of the hyperbola. The foci (focus F1 and focus F2) lie on the transverse line. The vertices are the points of intersection of both branches of the hyperbola with the transverse line. Both foci and vertices are symmetrical in relation to the conjugate axis, which implies that they'll have the same x-coordinate but with opposite signs.
The centre is the midpoint of the transverse axis and conjugate axis. This is where the two lines intersect.
Every hyperbola has two asymptotes (red dashed lines) that pass through the centre. As a hyperbola moves away from its centre, the branches approach these asymptotes. By definition, the branches of the hyperbola will never intersect it's asymptotes.
The central rectangle (orange dashed lines) is centred at the origin. The sides pass through each vertex and co-vertex.
It is helpful to identify this when graphing the hyperbola and its asymptotes. To draw the asymptotes of the hyperbola, all we have to do is extend the diagonals of the central rectangle.
Let us now derive the equation of a hyperbola centred at the origin.
Let P = (x, y), and the foci of a hyperbola centred at the origin be F1 = (–c, 0) and F2 = (c, 0)
From the plot above, (a, 0) is a vertex of the hyperbola, and so the distance from (−c, 0) to is . Similarly, the distance from (c, 0) to (a, 0) is .
The sum of the distances from the foci to the vertex is Let P(x, y) be a point on the hyperbola we want to investigate. From here, we can define d1 and d2 by:
d1 = distance from (c, 0) to (x, y)
d2 = distance from (–c, 0) to (x, y)
Squaring both sides, we obtain
Expanding the binomials and cancelling like terms,
Now dividing both sides by 4 and squaring both sides,
Expanding this and cancelling like terms again yields,
This simplifies to
Now dividing both sides by a2b2, the equation for the hyperbola becomes
as required! Below is a worked example that demonstrates the use of the Distance Formula in regards to hyperbolas.
Determine the equation of the hyperbola represented by the graph below.
Solution
The hyperbola below has foci at (0 , –5) and (0, 5) while the vertices are located at (0, –4) and (0, 4). The distance between these two coordinates is 8 units.
Thus, the difference between the distance from any point (x, y) on the hyperbola to the foci is 8 or –8 units, depending on the order in which you subtract.
Using the Distance Formula, we obtain the equation of the hyperbola as follows.
Let,
d1 = distance from (0, 5) to (x, y)
d2 = distance from (0, –5) to (x, y)
By dividing the expression by our final yield becomes
Let us now move on to the properties of hyperbolas. There are two cases to consider here:
Property | Centre (0,0) | |
Standard Form of Equation | ||
General Plot | ||
Opening | Opens left and right | Opens up and down |
Direction of Transverse Axis | Horizontal | Vertical |
Foci | (c, 0), (-c, 0) | (0, c), (0, -c) |
Vertices | (a, 0), (-a, 0) | (0, a), (0, -a) |
Length of Transverse Axis | 2a units | 2a units |
Length of Conjugate Axis | 2b unites | 2b units |
Equation of Asymptotes |
Property | Centre (h,k) | |
Standard Form of Equation | ||
General Plot | ||
Vertices | (h + a, k), (h - a, k) | (h, k + a), (h, k - a) |
Foci | (h + c, k), (h - c, k) | (h, k + c), (h, k - c) |
Equation of Asymptotes |
Identify the foci and vertices for the hyperbola .
Solution
The equation is of the form
Thus, the transverse axis lies on the y-axis. The centre is at the origin, so the y-intercepts are the vertices of the graph. We can thus find the vertices by setting x = 0 and solve for y as below.
The foci are located at and by the relationship between a, b and c established before, we obtain
Thus, the vertices are (0, –7) and (0, 7) while the foci are (0, –9) and (0, 9). The graph is shown below.
Identify the foci and vertices for the hyperbola .
Solution
The equation is of the form
Thus, the transverse axis lies on the y-axis. Here, h = 3 and k = 2, so the centre is at (3, 2). To find the vertices, we shall make use of the formula below.
The foci are located at and using as before, we obtain
Thus, the vertices are (3, –1) and (3, 5) while the foci are (0, –5.83) and (0, 5.83). The graph is shown below.
The equation for the asymptotes can be found using the formula given in the table. Give it a try for these examples! It is always helpful to sketch the asymptotes first before drawing the two branches of a hyperbola.
Express the following hyperbola in standard form given the following foci and vertices.
Solution
Notice that the vertices and foci are on the x-axis. Hence, the equation of the hyperbola will take the form
Since the vertices are then
Given the foci are then
Solving for b2 we obtain
Now that we have found a2 and b2, we can substitute this into the standard form as
The graph is shown below.
Express the following hyperbola in standard form given the following foci and vertices.
Solution
The vertices and foci have the same x-coordinates, so the transverse axis is parallel to the y-axis. The equation of the hyperbola will thus take the form
We must first identify the centre using the midpoint formula. The centre lies between the vertices (1, –2) and (1, 8), so
The length of the transverse axis, 2a, is bounded by the vertices. To find a2 we must evaluate the distance between the y-coordinates of the vertices.
The coordinates of the foci are
Using k + 3 = 16 and substituting k = 3, we obtain
Thus, we can solve for b2 by
Finally, substituting these values into the standard form, we obtain
The graph is shown below.
In this final section, we shall graph hyperbolas using the concepts introduced throughout this lesson.
Let us return to our previous examples for this segment,
Graph the hyperbola
Solution
As you can see, the hyperbola is already in the standard form.
This means that we have a pair of curves opening from the left and right. The vertices are and the foci are . The centre of the hyperbola is the origin, (0, 0). Here,
The equation of the asymptotes are
Always sketch the asymptotes when graphing hyperbolas. That way, we can accurately draw the curves associated with the equation.
The graph of is shown below.
In this section, it may be helpful to recall the method of Completing Squares to solve such problems.
Graph the hyperbola below.
Solution
To solve this expression, we must attempt to rearrange this in the standard form of a hyperbola. We do this by completing the square as follows.
We need to identify A and B. In doing so, we obtain
Dividing both sides by 225, we obtain the equation
The centre is (2, –1). We also have the values . Thus we obtain the following values for the vertices, foci and asymptotes.
The vertices
The foci
The asymptotes
The graph of is shown below.
The eccentricity of a conic section describes how closely related the curve is compared to a circle. The eccentricity is described by the variable e.
The eccentricity of a circle is zero, e = 0.
The eccentricity of a hyperbola is always more than 1, e > 1. The formula for finding the eccentricity of a hyperbola is given below.
The bigger the eccentricity, the less curved the conic section is.
Find the eccentricity of the hyperbola
Solution
Here, a2 = 25 and b2 = 9. Thus, the eccentricity of this hyperbola is given by
A hyperbola is the set of all points in a plane where the absolute value of the difference in distance between two fixed points, called the foci, is constant
The asymptotes for the general form of a hyperbola is given by y = ± bx/a
A hyperbola is a type of conic section made up of two parabola-like shaped curves. These pairs of curves can either open up and down or left and right. Additionally, each curve contains a vertex. However, it is crucial to note that these curves are not exactly parabolas (they only resemble them!)
What is the general form of a hyperbola centred at the origin?
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] or \[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \].
What is the general form of a hyperbola centered at \((h, k)\)?
\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] or \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \].
How do you describe a hyperbola?
A hyperbola is the set of all points in a plane where the absolute value of the difference in distance between the foci is constant.
How many axes of symmetry does a hyperbola have? What are their names?
A hyperbola has two axes of symmetry: the transverse axis and the conjugate axis.
What is the transverse axis?
The transverse axis is a line that passes through the centre of the hyperbola. The foci lie on this line and have vertices as their endpoints.
What is the conjugate axis?
The conjugate axis is a line perpendicular to the transverse axis and has co-vertices as its endpoints.
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