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Chapter 7: Problem 23

In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Foci \((0,\pm 4),\) one vertex (0,-2)

Short Answer

Expert verified
The equation of the hyperbola is \(\frac{y^2}{4} - \frac{x^2}{12} = 1\).

Step by step solution

01

Identify the orientation of the hyperbola

Since the foci are given as \((0, \pm 4)\), the hyperbola is centered on the y-axis and is vertically oriented. This means the general equation for the hyperbola will be of the form \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\).
02

Determine the center of the hyperbola

The center \((h, k)\) of the hyperbola is at the midpoint of the segment defined by the foci \((0,4)\) and \((0,-4)\). Thus, the center is \((0,0)\).
03

Identify the length of the semi-major axis \(a\)

Given that one vertex is at \((0, -2)\), the distance from the center of the hyperbola \((0,0)\) to the vertex is the length of \(a\), which is 2. Thus, \(a = 2\).
04

Calculate the distance between the foci \(c\)

The distance from the center \((0,0)\) to the foci \((0, \pm 4)\) is 4. So \(c = 4\).
05

Use the relationship between \(a\), \(b\), and \(c\)

For a hyperbola, the relationship \(c^2 = a^2 + b^2\) holds true. Substituting the known values, we have \(4^2 = 2^2 + b^2\), which simplifies to \(16 = 4 + b^2\).
06

Solve for \(b^2\)

From the equation \(16 = 4 + b^2\), we solve for \(b^2\) by subtracting 4 from both sides: \(b^2 = 12\).
07

Write the equation of the hyperbola

Substitute the values of the center \((0,0)\), \(a^2 = 4\), and \(b^2 = 12\) into the general form of the hyperbola equation: \(\frac{y^2}{4} - \frac{x^2}{12} = 1\).

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Key Concepts

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

Center of Hyperbola
The center of a hyperbola is the point around which the hyperbola is symmetrically organized. In our example, we are given foci at
  • (0, 4) and (0, -4)
, which tells us the line connecting these points lies on the y-axis. To find the center, you locate the midpoint between the two foci. The midpoint formula says to average the x-coordinates and the y-coordinates separately. Since both x-coordinates of the foci are 0 and the y-coordinates are 4 and -4 respectively, we calculate:\[\left(\frac{0+0}{2}, \frac{4 + (-4)}{2}\right)\]Simplifying gives us the center
  • (0, 0)
, meaning the entire hyperbola is positioned symmetrically around this point.
Semi-Major Axis
The semi-major axis of a hyperbola refers to the longest radius of an ellipse-like shape from its center to its vertex. In a vertically oriented hyperbola, this runs along the y-direction. In our specific problem, one vertex is at (0, -2). Since the hyperbola's center is at (0, 0), the vertical distance from the center to this vertex is 2. Hence, for this hyperbola, we set the semi-major axis
  • a = 2
. The semi-major axis helps to determine how widely the hyperbola opens along the y-axis for our example.
Distance Between the Foci
The distance between the foci of a hyperbola, symbolized as c, is significant as it determines the stretch of the hyperbola along its primary axis. For the given system, the foci are at
  • (0, 4) and (0, -4)
. We calculate the distance between these points using the formula for distance between two points, which gives:\[|4 - (-4)| = 8\]However, since we deal with the center of the hyperbola, we use only half of this distance for c, which is
  • c = 4
. This distance affects the slope of the hyperbola's parabola-like branches.
Relationship Between a, b, and c
For hyperbolas, the values of \(a\), \(b\), and \(c\) are connected by the equation:\[c^2 = a^2 + b^2\]In our problem, we know
  • a = 2
  • c = 4
, which allows us to plug these into the relationship:\[4^2 = 2^2 + b^2\]This solves as:\[16 = 4 + b^2\]By subtracting 4 from both sides, we find:
  • b^2 = 12
. This relationship is crucial for deriving the exact shape of the hyperbola and understanding how its axes compare in length. With this, you can specify the full equation of the hyperbola and its behavior in a coordinate plane.

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Most popular questions from this chapter

In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 $$

Conjugate Hyperbolas The hyperbolas $$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \quad \text { and } \quad \frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1 $$ are said to be conjugates of each other. (a) Find an equation of the hyperbola that is conjugate to $$ \frac{x^{2}}{25}-\frac{y^{2}}{144}=1 $$ (b) Discuss how the graphs of conjugate hyperbolas are related.

Find the center, foci, vertices, endpoints of the minor axis, and eccentricity of the given ellipse. Graph the ellipse. $$ 4 x^{2}+\left(y+\frac{1}{2}\right)^{2}=4 $$

In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Foci \((\pm 4,0),\) length of transverse axis 6

Three points are located at \(A(-10,16), B(-2,0),\) and \(C(2,0),\) where the units are kilometers. An artillery gun is known to lie on the line segment between \(A\) and \(C,\) and using sounding techniques it is determined that the gun is \(2 \mathrm{~km}\) closer to \(B\) than to \(C\). Find the point where the gun is located.
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