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Chapter 7: Problem 70

Find an equation for each hyperbola. Vertices \((0,5)\) and \((0,-5) ;\) passing through \((3,10)\)

Short Answer

Expert verified
The equation is \( \frac{y^2}{25} - \frac{x^2}{3} = 1 \).

Step by step solution

01

Identify the Center from Vertices

The vertices (0,5) and (0,-5) give us information about the position and size of the hyperbola. Since both vertices lie on the y-axis, we know the hyperbola is vertical. The center of the hyperbola is the midpoint of (0,5) and (0,-5), which is (0,0).
02

Determine the Parameter a

The distance between the center (0,0) and either vertex (0,5) or (0,-5) is 5. In a hyperbola, this distance is represented by the parameter 'a'. Therefore, a = 5.
03

Write the General Equation of the Hyperbola

For a vertical hyperbola centered at (0,0), the standard form of the equation is: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]Substituting a = 5, we have: \[ \frac{y^2}{25} - \frac{x^2}{b^2} = 1 \]
04

Use the Given Point to Solve for b

The hyperbola also passes through the point (3,10). Substitute x=3, y=10 into the equation: \[ \frac{10^2}{25} - \frac{3^2}{b^2} = 1 \]\[ \frac{100}{25} - \frac{9}{b^2} = 1 \] \[ 4 - \frac{9}{b^2} = 1 \] \[ \frac{9}{b^2} = 3 \] \[ b^2 = 3 \] Thus, b is the square root of 3.
05

Write the Final Equation of the Hyperbola

Now that we have determined b^2 = 3, substitute this back into the hyperbola equation: \[ \frac{y^2}{25} - \frac{x^2}{3} = 1 \] This is the equation of the hyperbola.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertices
Vertices are two specific points on a hyperbola, situated along its axis. They are critical in determining the shape and orientation of the hyperbola.
Understanding the location of vertices helps in sketching the structure of the hyperbola and provides insight into its dimensions.

Here, the vertices are \( (0,5) \) and \( (0,-5) \), indicating a vertical hyperbola.
  • The vertices lie along the y-axis, confirming the orientation.
  • From the vertices, you can determine the direction: vertical hyperbolas open upward and downward.
Since these points are equidistant from the center of the hyperbola, the midpoint can be easily calculated, aiding in locating the center.
Center of Hyperbola
The center of a hyperbola is the midpoint between its vertices. It acts as the equilibrium point from which the hyperbola stretches out in both directions. In geometric terms, the center is essentially the average point between the vertices.

In the given exercise, the vertices \( (0,5) \) and \( (0,-5) \) pin the center down at \( (0,0) \).
  • The center is the starting point for both the transverse and conjugate axes.
  • It serves as the origin for measuring distances such as 'a' or 'b'.
Having the center readily identified makes it easier to apply the standard form equation of the hyperbola. Knowing the center helps in understanding and defining both axes clearly.
Standard Form of Hyperbola
The standard form of a hyperbola's equation is crucial for expressing its characteristics mathematically. A vertical hyperbola centered at the origin has its equation defined as:

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

This equation's structure captures how the y-term and x-term relate to the parameters 'a' and 'b' respectively.
  • 'a' relates to the distance from the center to the vertices along the vertical axis.
  • 'b' describes the distance from the center along the horizontal axis, though it's not the foremost determining feature for the hyperbola's opening.
This structure is indispensable because it succinctly characterizes a hyperbola's spread based on its centered origin and its relationship with 'a' and 'b' parameters.
Parameter a
Within a hyperbola, the parameter 'a' signifies the distance from the center to each vertex along the transverse axis. It essentially controls how the hyperbola stretches perpendicular to the conjugate axis.

Using the exercise example, the given vertices \( (0,5) \) and \( (0,-5) \) let us determine that:\( a = 5 \).
  • 'a' is always a positive value as it denotes span, regardless of direction.
  • In the equation, it determines the denominator under the y-term for vertical hyperbolas.
Understanding 'a' is pivotal because it provides the precise measurement needed to anchor the hyperbola's equation. Calculating 'a' correctly ensures that the opening of the hyperbola aligns accurately with the features specified in the problem.

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Most popular questions from this chapter

Solve each problem. The orbit of Venus is an ellipse, with the sun at one focus. An approximate equation for the orbit is $$\frac{x^{2}}{5013}+\frac{y^{2}}{4970}=1$$ where \(x\) and \(y\) are measured in millions of miles. (a) Approximate the length of the major axis. (b) Approximate the length of the minor axis.

Find an equation for each ellipse. Center \((-3,6) ;\) major axis vertical, with length \(10 ; c=2\)

Find an equation for each ellipse. Center \((2,0)\); minor axis of length 6 ; major axis horizontal and of length 9

Find an equation for each ellipse. Center \((3,-2) ; a=5 ; c=3 ;\) major axis vertical

Solve each problem. Structure of an Atom In \(1911,\) Ernest Rutherford discovered the basic structure of the atom by "shooting" positively charged alpha particles with a speed of \(10^{7}\) meters per second at a piece of gold foil \(6 \times 10^{-7}\) meter thick. Only a small percentage of the alpha particles struck a gold nucleus head-on and were deflected directly back toward their source. The rest of the particles often followed a hyperbolic trajectory because they were repelled by positively charged gold nuclei. Thus, Rutherford proposed that the atom was composed of mostly empty space and a small, dense nucleus. The figure shows an alpha particle \(A\) initially approaching a gold nucleus \(N\) and being deflected at an angle \(\theta=90^{\circ}\) \(N\) is located at a focus of the hyperbola, and the trajectory of \(A\) passes through a vertex of the hyperbola. (a) Determine the equation of the trajectory of the alpha particle if \(d=5 \times 10^{-14}\) meter. (b) Approximate the minimum distance between the centers of the alpha particle and the gold nucleus. (GRAPH CAN'T COPY)
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