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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 10: Q. 48 (page 668)

Find an equation for the hyperbola described. Graph the equation by hand.
Vertices at $(1, - 3)$ and $(1, 1)$ ; asymptote the line $y + 1 = \frac{3}{2} (x - 1)$

Short Answer

Expert verified

The hyperbola is

$\frac{{(y + 1)}^{2}}{9} - \frac{{(x - 1)}^{2}}{4} = 1$

The graph of the hyperbola is

Step by step solution

01

Given data

Vertices are at $(1, - 3)$ and $(1, 1)$ and the asymptote line is $y + 1 = \frac{3}{2} (x - 1)$ .

02

Concept used

The hyperbola with center $(h, k)$ and transverse line parallel to role="math" localid="1647059465861" $y$ -axis is given by

$\frac{{(y - k)}^{2}}{a^{2}} - \frac{{(x - h)}^{2}}{b^{2}} = 1$ ; $b^{2} = c^{2} - a^{2}$

where focus are at $(h, k 卤 c)$ , vertices are at $(h, k 卤 a)$ and the asymptotes are $y - k = 卤 \frac{a}{b} (x - h)$

03

Application

Since the vertices are lie on the line $x = 1$ , the transverse axis is parallel to the $y$ -axis.

So, $h = 1$

Given that, vertices are $(1, - 3), (1, 1)$ and one asymptote is $y + 1 = \frac{3}{2} (x - 1)$ .

Since, in general form of asymptotes are $y - k = 卤 \frac{a}{b} (x - h)$

So, $h = 1$

$k = - 1$

$a = 3$

$b = 2$

Thus, the equation of hyperbola is

$\frac{{(y - k)}^{2}}{a^{2}} - \frac{{(x - h)}^{2}}{b^{2}} = 1$

Substituting all values,

$\frac{{(y + 1)}^{2}}{3^{3}} - \frac{{(x - 1)}^{2}}{2^{2}} = 1$

So, role="math" localid="1647060214960" $\frac{{(y + 1)}^{2}}{9} - \frac{{(x - 1)}^{2}}{4} = 1$

04

Graph

The hyperbola is

$\frac{{(y + 1)}^{2}}{9} - \frac{{(x - 1)}^{2}}{4} = 1$

The graph will be

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