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Chapter 12: Problem 41

Find an equation of the following ellipses and hyperbolas, assuming the center is at the origin. A hyperbola with vertices (卤4,0) and foci (卤6,0)

Short Answer

Expert verified
Answer: The equation of the hyperbola is given by: \[\frac{x^2}{16} - \frac{y^2}{20} = 1\]

Step by step solution

01

Determine the semi-major axis, a

From the given vertices (卤4, 0), we can observe that the distance from the center to the vertices (along the x-axis) is 4. This distance is the same as the semi-major axis, a. Therefore, a = 4.
02

Determine the distance between a focus and the center, c

From the given foci (卤6, 0), we can observe that the distance from the center to the foci (along the x-axis) is 6. This distance is the same as the value c. Therefore, c = 6.
03

Calculate the semi-minor axis, b

Now, we will use the relationship between a, b, and c in order to find the semi-minor axis, b. The relationship is given by: \[c^2 = a^2 + b^2\] Substitute the values of a and c: \[6^2 = 4^2 + b^2\] Solve for b^2: \[b^2 = 36 - 16 = 20\]
04

Write the equation of the hyperbola

Now that we have the values of a^2 and b^2, we can write the equation of the hyperbola using the standard formula. \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\] Substitute the values of a^2 and b^2: \[\frac{x^2}{(4)^2} - \frac{y^2}{20} = 1\] So, the equation of the hyperbola is: \[\frac{x^2}{16} - \frac{y^2}{20} = 1\]

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Key Concepts

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

Vertices and Foci
When dealing with hyperbolas, the vertices and foci are crucial markers that help define the shape and position of the hyperbola. The vertices of a hyperbola represent the closest points the hyperbola reaches to the center. In this exercise, the vertices are located at \((\pm 4, 0)\), indicating that they lie on the x-axis. Measure these distances from the center point \,\( (0,0) \,\), since that's where the hyperbola is centered in this scenario.

The foci are points located inside each branch of the hyperbola. They are key in defining the "spread" or openness of the hyperbola. For this hyperbola, the foci are at \((\pm 6, 0)\). They too are aligned along the x-axis as this is a horizontally oriented hyperbola. The distance from the center to each focus is denoted by \(c\), and in this example, \(c = 6\).

  • The distance from the center to each vertex is \(a = 4\).
  • The distance from the center to each focus is \(c = 6\).
Semi-Major and Semi-Minor Axis
In hyperbolas, unlike ellipses, the terms "semi-major" and "semi-minor" axis refer to different concepts. They help us understand the dimensions and size of the hyperbola. For this exercise, if the vertices are \((\pm 4, 0)\), the semi-major axis is horizontal, meaning the hyperbola opens left and right. The semi-major axis has a length of \(2a\), with each leg extending \(a = 4\) units from the center. Thus, its length is \(8\) units in total.

On the other hand, the semi-minor axis expresses the vertical "spread" between the two branches of the hyperbola. Its length will determine how close or wide the hyperbola's branches are from each other vertically. This is calculated using the relation \(c^2 = a^2 + b^2\).

With \(c = 6\) and \(a = 4\), we find \(b\) by solving:\[b^2 = c^2 - a^2 = 36 - 16 = 20\]Therefore, \(b \approx \sqrt{20}\).

  • Length of the semi-major axis: \(8\) units.
  • Length of the semi-minor axis: \(2\sqrt{20}\) units vertically.
Standard Form of Hyperbola Equation
Every hyperbola equation can be expressed in a standard form that helps easily recognize its structural properties. The general equation of a hyperbola centered at the origin with horizontal transverse axis is:\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]
For our specific scenario, with \(a = 4\) and \(b^2 = 20\), we substitute these values into the formula. The equation of the hyperbola becomes:\[\frac{x^2}{16} - \frac{y^2}{20} = 1\]

Here's how the components break down:
  • \( \frac{x^2}{16} \): This denotes the horizontal opening, influenced by the semi-major axis \(a\).
  • \( \frac{y^2}{20} \): This represents the vertical component, tied to the semi-minor axis \(b\).
  • The equation set equal to \(1\) signifies the standard nature of the hyperbola's form.
Understanding this formula allows for quick identification of the hyperbola's orientation and dimensions based on its symmetry and axes.

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Most popular questions from this chapter

Consider the region \(R\) bounded by the right branch of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and the vertical line through the right focus. a. What is the volume of the solid that is generated when \(R\) is revolved about the \(x\) -axis? b. What is the volume of the solid that is generated when \(R\) is revolved about the \(y\) -axis?

Find the arc length of the following curves on the given interval. $$x=t^{4}, y=\frac{t^{6}}{3} ; 0 \leq t \leq 1$$

Suppose the function \(y=h(x)\) is nonnegative and continuous on \([\alpha, \beta],\) which implies that the area bounded by the graph of h and the x-axis on \([\alpha, \beta]\) equals \(\int_{\alpha}^{\beta} h(x) d x\) or \(\int_{\alpha}^{\beta} y d x .\) If the graph of \(y=h(x)\) on \([\alpha, \beta]\) is traced exactly once by the parametric equations \(x=f(t), y=g(t),\) for \(a \leq t \leq b,\) then it follows by substitution that the area bounded by h is $$\begin{array}{l}\int_{\alpha}^{\beta} h(x) d x=\int_{\alpha}^{\beta} y d x=\int_{a}^{b} g(t) f^{\prime}(t) d t \text { if } \alpha=f(a) \text { and } \beta=f(b) \\\\\left(\text { or } \int_{\alpha}^{\beta} h(x) d x=\int_{b}^{a} g(t) f^{\prime}(t) d t \text { if } \alpha=f(b) \text { and } \beta=f(a)\right)\end{array}$$. Find the area of the region bounded by the astroid \(x=\cos ^{3} t, y=\sin ^{3} t,\) for \(0 \leq t \leq 2 \pi\) (see Example 8 Figure 12.17 ).

Determine whether the following statements are true and give an explanation or counterexample. a. The equations \(x=-\cos t, y=-\sin t,\) for \(0 \leq t \leq 2 \pi\) generate a circle in the clockwise direction. b. An object following the parametric curve \(x=2 \cos 2 \pi t\) \(y=2 \sin 2 \pi t\) circles the origin once every 1 time unit. c. The parametric equations \(x=t, y=t^{2},\) for \(t \geq 0,\) describe the complete parabola \(y=x^{2}\) d. The parametric equations \(x=\cos t, y=\sin t,\) for \(-\pi / 2 \leq t \leq \pi / 2,\) describe a semicircle. e. There are two points on the curve \(x=-4 \cos t, y=\sin t,\) for \(0 \leq t \leq 2 \pi,\) at which there is a vertical tangent line.

Consider an ellipse to be the set of points in a plane whose distances from two fixed points have a constant sum \(2 a\). Derive the equation of an ellipse. Assume the two fixed points are on the \(x\) -axis equidistant from the origin.
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