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Chapter 9: Problem 6

Find the standard form of the equation of each hyperbola satisfying the given conditions. $$\text { Foci: }(0,-6),(0,6) ; \text { vertices: }(0,-2),(0,2)$$

Short Answer

Expert verified
The standard form of the equation of the hyperbola is \(x^2/4 - y^2/32 = 1\).

Step by step solution

01

Identify The Type of Hyperbola

From the given vertices and foci, it is evident that both lie along the y-axis. When the vertices and foci are along the y-axis, the hyperbola opens either up or down - in this case, it's a vertical hyperbola.
02

Determine the Center, a and c

The center of the hyperbola is the midpoint between the two vertices or the two foci. All of which lie at the origin (0,0). Vertices are a units from the center, thus the value of |a| is the distance from the center to a vertex, which is 2. The foci are c units from the center, thus the value of |c| is the distance from the center to a focus, which is 6.
03

Calculate b

Now calculate b using the formula \(c^2 = a^2 + b^2\). Simplifying, it will be \(b = \sqrt{c^2 - a^2} = \sqrt{6^2 - 2^2} = \sqrt{32}\).
04

Write The Equation in Standard Form

Now that we have the center (h,k), a, b for a vertical-opening hyperbola, we can plug them into the standard equation \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), which provides the final equation as \(x^2/4 - y^2/32 = 1\).

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Key Concepts

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

Standard Form of Hyperbola
The standard form of a hyperbola's equation is crucial for understanding its structure. Hyperbolas can open either horizontally or vertically. For a hyperbola that opens vertically, the center is at \(h, k\), and its standard form is given by the equation: \[(y - k)^2/a^2 - (x - h)^2/b^2 = 1\].

This form indicates that the hyperbola opens along the y-axis. The values of \(a^2\) and \(b^2\) determine the spread of the hyperbola along the corresponding axes.
  • In a vertical hyperbola, \(a\) relates to the distance to the vertices along the y-axis.
  • \(b\) provides the extent of distance from the center to the hyperbola's co-vertices on the x-axis.
Remember that the standard form makes it easy to identify key characteristics like the center, vertices, and other properties at a glance.
Foci and Vertices of Hyperbola
The placement of foci and vertices provides significant insight into a hyperbola's orientation and dimensions. Foci are always located along the axis the hyperbola opens. For vertical hyperbolas:
  • The distance |\(c\)| from the center to each focus extends along the vertical y-axis.
  • Vertices are \(b\) distance away along the x-axis for a vertically-opening hyperbola, but with vertical focus, \(|a|\) measures from the center to the vertices, along the y-axis.
For the problem at hand, the center \(0,0\) has foci at \(0,-6\) and \(0,6\), and the vertices at \(0,-2\) and \(0,2\).

This setup clearly defines a vertically oriented hyperbola with a centered origin.
Equation of Hyperbola
Equations of hyperbolas can vary in complexity with respect to the values of \(a\), \(b\), and \(c\). To construct a hyperbola equation:
  1. Start with the formula representing its orientation. For vertical: \[(y - k)^2/a^2 - (x - h)^2/b^2 = 1\].
  2. Substitute defined values into the equation. Solving the given exercise, \(a \) is calculated as 2 from vertex distances, \(c \) is 6 from focal distances.
  3. Calculate \(b \) using \(c^2 = a^2 + b^2\).
The computation in Turn gives \((x+0)^2/4 - (y+0)^2/32 = 1\).

The subtraction of terms in the equation defines it as a hyperbola, and this particular setup points to a vertical orientation.
Vertical Hyperbola Calculus
Calculus allows us to explore hyperbolas further, especially when differentiating their equations. Knowing we have a vertical hyperbola in place, its derivative provides insight into the hyperbola's slope at various points.
  • The slope of the tangent can be found by differentiating its equation with respect to x or y.
  • For a vertical hyperbola, expect symmetric behavior along the y-axis.


Visualizing calculation connections can also involve analyzing surface areas or other integral measures of the hyperbola. However, remember the main difference: vertical hyperbolas primarily shift along the y-axis, hence, direct focus is given to y-related differentials.

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Most popular questions from this chapter

In Exercises \(9-20,\) use point plotting to graph the plane curve described by the given parametric equations. Use arrows to show the orientation of the curve corresponding to increasing values of \(t\). $$x=|t+1|, y=t-2 ;-\infty < t < \infty$$

Identify the conic and graph the equation: $$r=\frac{4 \sec \theta}{2 \sec \theta-1}$$

Will help you prepare for the material covered in the first section of the next chapter. Evaluate \(\frac{(-1)^{n}}{3^{n}-1}\) for \(n=1,2,3,\) and 4

Eliminate the parameter: \(x=\cos ^{3} t\) and \(y=\sin ^{3} t\)

The plane curve described by the parametric equations \(x=3 \cos t \quad\) and \(\quad y=3 \sin t, \quad 0 \leq t<2 \pi, \quad\) has \(\quad\) a counterclockwise orientation. Alter one or both parametric equations so that you obtain the same plane curve with the opposite orientation.
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