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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. 47 (page 668)

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

Short Answer

Expert verified

The equation of the hyperbola is

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

The graph of the hyperbola is

Step by step solution

01

Given data

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

02

Concept used

The hyperbola with center $(h, k)$ and transverse axis parallel to $x$ -axis is given by

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

where focus is at role="math" localid="1647055818744" $(h 卤 c, k)$ , vertices are at $(h 卤 a, k)$ and asymptotes are $y - k = 卤 \frac{b}{a} (x - h)$ .

03

Application

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

So, $k = - 1$

Given that the vertices are $(- 1, - 1)$ and $(3, - 1)$

Since asymptotes are $y - k = 卤 \frac{b}{a} (x - a)$ ,

Also given that, the one asymptote line is

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

So, $h = 1$

$k = - 1$

$a = 2$

$b = 3$

Thus, the equation of the hyperbola is

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

Substituting all values,

role="math" localid="1647057102885" $\frac{{(x - 1)}^{2}}{2^{2}} - \frac{{(y + 1)}^{2}}{3^{2}} = 1$

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

04

Graph

The hyperbola is,

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

The graph will be

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Most popular questions from this chapter

In Problems 31鈥� 42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21鈥�30 for Problems 31鈥� 40. $13 x^{2} - 6 \sqrt{3} x y + 7 y^{2} - 16 = 0$

In Problems 63鈥� 66, graph each function by hand. Be sure to label any intercepts

$f (x) = \sqrt{- 1 + x^{2}}$

Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation by hand.

Focus is at $(0, - 3)$ and vertex at $(0, 0)$ .

True or False

The equation $3 x^{2} + B x y + 12 y^{2} = 10$ defines a parabola if $B = - 12$ .

In Problems 43鈥�52, identify the graph of each equation without applying a rotation of axes.

2 x^{2} - 3 x y + 2 y^{2} - 4 x - 2 = 0

See all solutions

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