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Chapter 6: Problem 23

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$\frac{y^{2}}{1}-\frac{x^{2}}{4}=1$$

Short Answer

Expert verified
The center of the hyperbola is at the origin (0,0), vertices are at (0,1) and (0,-1), foci are at (0,\(\sqrt{5}\)) and (0,-\(\sqrt{5}\)), and the equations of the asymptotes are \(y = 2x\) and \(y = -2x\).

Step by step solution

01

Identify the Form

The given equation is \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\) which is in the standard form of a vertical hyperbola. Hence, it is determined that \(b^2 = 1\) and \(a^2 = 4\). With this information, find the values of \(a\) and \(b\).
02

Compute a and b

By comparing the equation to the standard form, we find that \(a = \sqrt{4} = 2\) and \(b = \sqrt{1} = 1\).
03

Compute c

The distance to the foci from the center is \(c = \sqrt{a^{2} + b^{2}}\). We substitute the values of \(a\) and \(b\) and find that \(c = \sqrt{2^{2} + 1^{2}} = \sqrt{5}\). The foci are located at coordinates \((0, \pm \sqrt{5})\).
04

Find Asymptote equations

The equation for the asymptotes for a vertical hyperbola is \(x = \pm \frac{by}{a}\). Substituting \(b = 1\) and \(a = 2\) we get the equations of the asymptotes as \(x = \pm \frac{y}{2}\) or \(y = \pm 2x\).
05

Sketch the Hyperbola and Asymptotes

Firstly, draw the asymptotes whose equations were found in the previous step. Then, draw the hyperbola opening upwards and downwards with respect to the asymptotes. The vertices are at \((0,\pm1)\) and the foci are at \((0,\pm\sqrt{5})\). Thus, a sketch of the hyperbola should reflect these characteristics.

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