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Chapter 6: Problem 21

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$\frac{y^{2}}{25}-\frac{x^{2}}{81}=1$$

Short Answer

Expert verified
The center of the hyperbola is at the origin (0, 0). The vertices are at (0, 卤5). The foci are at (0, 卤\sqrt{106}). The equations of the asymptotes are \(y = \pm \frac{5}{9}x\). The sketch should show an ever-widening curve approaching the asymptotes.

Step by step solution

01

Finding the center

The center of the hyperbola is at the origin (0, 0) because there is no constant term that can shift the hyperbola from the origin.
02

Finding the vertices

The vertices of the hyperbola are found from the \(a\) term in the equation, which is \(a^2 = 25\). So, \(a = \pm5\). Thus, the vertices are at (0, 卤5).
03

Finding the foci

The foci are found by solving for \(c\), where \(c^2 = a^2 + b^2\). Here, \(b^2 = 81\), so \(b = 9\). So, \(c^2 = 25 + 81 = 106\), and \(c = \sqrt{106}\). Thus, the foci are at (0, 卤\sqrt{106}).
04

Finding the equations of the asymptotes

The equations of the asymptotes of the hyperbola are given by \(y = \pm \frac{a}{b} x = \pm \frac{5}{9}x\). Thus, the asymptotes are lines passing through the origin with slopes of 卤5/9.
05

Sketching the hyperbola

Start by plotting the center, vertices and foci on the coordinate plane. Then, draw the lines representing the asymptotes. The hyperbola should be drawn as an ever-widening curve that approaches asymptotes as \(x \rightarrow \pm \infty\).

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