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Chapter 10: Problem 44

Find the vertices and the asymptotes of each hyperbola. $$ 9 y^{2}-36 x^{2}=144 $$

Short Answer

Expert verified
Vertices are at (0, -4) and (0, 4) and Asymptotes have equations \(y = 2x\) and \(y = -2x\).

Step by step solution

01

Get the standard form of the equation

To get the standard form of the equation, divide the given equation by 144. So, the equation becomes \(\frac{y^2}{16} - \frac{x^2}{4} = 1\).
02

Identify the center, a and b

From the standard form, we see that the center of the hyperbola is at the origin, (0,0), since h and k are both 0. The value of \(a\) is 4 and \(b\) is 2 because in the standard form of equation of hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), \(a^2\) equals 16 and \(b^2\) equals 4.
03

Find the vertices

The vertices of a hyperbola are a distance 'a' from the center. As the major axis is vertical in this case, the vertices are at (0,卤4).
04

Find the asymptotes

The equations of the asymptotes in this case are y = 卤(a/b)x. The equations of the asymptotes are therefore \(y = 卤2x\).

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Key Concepts

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

Standard Form of a Hyperbola
The standard form of a hyperbola helps to identify the shape and orientation of the curve. For any hyperbola, the equation can take two forms:
  • Horizontal hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
  • Vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)
The difference lies in which term comes first. If \(y\) is first, the hyperbola opens upward and downward, indicating a vertical hyperbola. When \(x\) is first, the hyperbola opens left and right, indicating a horizontal hyperbola. After transforming the equation \(9y^2 - 36x^2 = 144\) by dividing it by 144, we get \(\frac{y^2}{16} - \frac{x^2}{4} = 1\). This form tells us we are dealing with a vertical hyperbola.
Vertices of a Hyperbola
Vertices are key components of a hyperbola, representing the points where the curve is closest to its center. For a hyperbola given by the standard form, vertices are found at a distance \(a\) from the center along the major axis. In the equation \(\frac{y^2}{16} - \frac{x^2}{4} = 1\), we identify \(a^2\) as 16, thus \(a\) as 4. The center of our hyperbola is at \((0,0)\), leading us to find the vertices at \((0, +4)\) and \((0, -4)\). These vertices tell us that the hyperbola stretches vertically along the \(y\)-axis.
Asymptotes of a Hyperbola
Asymptotes are lines that a hyperbola approaches but never actually reaches. For hyperbolas, they provide a way to sketch the curve correctly by showing the shape's extent and direction. The equations for asymptotes in our vertical hyperbola \(\frac{y^2}{16} - \frac{x^2}{4} = 1\) are derived from the slope \(\pm \frac{a}{b}\). Substituting \(a = 4\) and \(b = 2\), we obtain the asymptotes as \(y = \pm 2x\). This means the hyperbola will stretch outwards to match the angle formed by these paths.
Conic Sections
Conic sections are curves obtained by intersecting a plane with a cone. Depending on the angle and position of the intersection, they can form different shapes such as circles, ellipses, parabolas, and hyperbolas. Hyperbolas arise when a plane slices through both halves of a double cone, creating two mirrored curves. They have two characteristic branches, which are symmetrical and extend indefinitely. Understanding the properties of a hyperbola, like its shape, vertices, and asymptotes, is essential to mastering conic sections. Each type of conic section has unique properties and applications in various fields, from architecture to astronomy.

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