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Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$9 x^{2}-16 y^{2}=1$$

Short Answer

Expert verified
Vertices: \(\left(\pm \frac{1}{3}, 0\right)\); Foci: \(\left(\pm \frac{5}{12}, 0\right)\); Asymptotes: \(y = \pm \frac{3}{4}x\).

Step by step solution

01

Identify the form of the equation

The given equation is \(9x^2 - 16y^2 = 1\). This is a standard form of a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) where the transverse axis is horizontal.
02

Rewrite the equation in standard form

We start by rewriting the given equation as fractions to identify values more easily: \(\frac{x^2}{\frac{1}{9}} - \frac{y^2}{\frac{1}{16}} = 1\). Now let's rewrite the denominators in terms of perfect squares: \(\frac{x^2}{\left(\frac{1}{3}\right)^2} - \frac{y^2}{\left(\frac{1}{4}\right)^2} = 1\). Thus, \(a = \frac{1}{3}\) and \(b = \frac{1}{4}\).
03

Determine the vertices

The vertices of the hyperbola are located at \( (\pm a, 0) \). With \(a = \frac{1}{3}\), the vertices are \( \left(\pm \frac{1}{3}, 0\right) \).
04

Determine the foci

For a hyperbola, the relationship \(c^2 = a^2 + b^2\) calculates the value \(c\). Here, \(a = \frac{1}{3}\) and \(b = \frac{1}{4}\), so: \(c = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{4}\right)^2} = \sqrt{\frac{1}{9} + \frac{1}{16}} = \sqrt{\frac{25}{144}} = \frac{5}{12}\). Thus, the foci are at \(\left(\pm \frac{5}{12}, 0\right)\).
05

Find the asymptotes

For hyperbolas in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the asymptotes are given by \(y = \pm \frac{b}{a}x\). So here, \(y = \pm \frac{\frac{1}{4}}{\frac{1}{3}}x = \pm \frac{3}{4}x\).
06

Sketch the graph

Plot the vertices \(\left(\pm \frac{1}{3}, 0\right)\) and the foci \(\left(\pm \frac{5}{12}, 0\right)\) on the graph. Draw the asymptotes through the origin with slopes \(y = \pm \frac{3}{4}x\). Then, sketch the two branches of the hyperbola opening horizontally around these asymptotes.

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Key Concepts

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

Vertices of a Hyperbola
In a hyperbola, the vertices are crucial points that lie on the hyperbola itself, representing the closest points to each other on either branch. Given the standard form of a hyperbola with a horizontal transverse axis, the equation can be written as:\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]Here, the vertices are located at \((\pm a, 0)\). For our specific hyperbola, the value of \(a\) is found by rewriting the equation and identifying the denominator under \(x^2\). From the steps provided, we know \(a = \frac{1}{3}\). Thus, the vertices are located at the points \(\left(\pm \frac{1}{3}, 0\right)\).
This means the vertices are slightly set apart on the horizontal axis, indicating the primary direction of the hyperbola's branches opening out.
Foci of a Hyperbola
The foci of a hyperbola are two fixed points located along the transverse axis, further apart than the vertices. These points are used to define the shape of the hyperbola.
In the context of a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the distance to the foci from the center of the hyperbola can be calculated with the formula:\[c^2 = a^2 + b^2\]Given \(a = \frac{1}{3}\) and \(b = \frac{1}{4}\), we calculate \(c\) as:\[c = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{4}\right)^2} = \sqrt{\frac{25}{144}} = \frac{5}{12}\]This tells us that the foci are located at \(\left(\pm \frac{5}{12}, 0\right)\). The foci are always further from the center than the vertices, which in this case are closer to the origin, mirroring the stronger curves on the graph.
Asymptotes of a Hyperbola
Asymptotes are lines that the hyperbola approaches but never actually meets. These lines help guide the shape and direction of the hyperbola's branches.
For a hyperbola with a standard form equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations for the asymptotes are determined by:\[y = \pm \frac{b}{a}x\]In our example, with \(a = \frac{1}{3}\) and \(b = \frac{1}{4}\), the slopes of the asymptotes are calculated to be \(\frac{b}{a} = \frac{1/4}{1/3} = \frac{3}{4}\).
Thus, the asymptotes are described by the lines:
  • \(y = \frac{3}{4}x\)
  • \(y = -\frac{3}{4}x\)
These lines intersect at the center of the hyperbola (origin) and provide a visual guide for the hyperbola's branches as they extend infinitely without touching the asymptotes.
Equation of a Hyperbola
The equation of a hyperbola is fundamental in defining its shape and properties. A hyperbola can be expressed in a couple of standard forms, but our specific scenario involves the form where the transverse axis is horizontal:\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]In this equation:
  • \(a^2\) is associated with the horizontal axis (\(x\)-axis).
  • \(b^2\) is associated with the vertical axis (\(y\)-axis).
Rewriting our initial hyperbola equation \(9x^2 - 16y^2 = 1\) into this form, we get:\[\frac{x^2}{\left(\frac{1}{3}\right)^2} - \frac{y^2}{\left(\frac{1}{4}\right)^2} = 1\]This rewritten equation clearly shows the relationship between \(x\) and \(y\), helping to understand which axis the hyperbola spreads across, showing a symmetric pattern with respect to the center. It is this equation that provides the foundation for plotting and analyzing the hyperbola.

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