Precalculus

9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Question

Chapter 10, Problem 3RE

To determine

Tofind:the equation and if the equation is parabolic give its vertex, focus and directrix; if its ellipse gives its center, vertices and foci; if it’s is a hyperbola, give its center, vertices, foci and asymptotes.

Expert Solution & Answer

Answer to Problem 3RE

The center= $(0,0)$ , Foci= $(±c,0)=(±26,0)$ , vertices is $(±a,0)=(±5,0)$ and Asymptotes: $y=±bax⇒y=±15x$ .

Explanation of Solution

Given:

$x225-y2=1$ .

Concept used:

Formula for hyperbola forming open left and right.

Form: $(x-h)2a2-(y-k)2b2=1$ .

Center= $(h,k)$ .

Vertices= $(h+a,k)$ and $(h-a,k)$ .

Slope of asymptotes: $±ba$ .

Equation of asymptotes: $y=k±ba(x-h)$ .

Formula for hyperbola opening up and down:

Form: $(y-k)2b2-(x-h)2a2=1$ .

Center: $(h,k)$ .

Vertices: $(h,k+b) and (h,k-b)$ .

Slope of asymptotes: $±ba$ .

Calculation:

$x225−y2=1$ .

Here the equation is in the form of $x2a2-y2b2=1$ where $a=5,b=1$ ,it’s a hyperbola

The center= $(0,0)$ The vertices is $(±a,0)=(±5,0)$ .

Foci= $(±c,0)=(±26,0)$ where $c=a2+b2=26.$

Asymptotes: $y=±bax⇒y=±15x$ .

Hence, The center= $(0,0)$ , Foci= $(±c,0)=(±26,0)$ , vertices is $(±a,0)=(±5,0)$ and Asymptotes: $y=±bax⇒y=±15x$ .

Chapter 10, Problem 2RE

Chapter 10, Problem 4RE

Chapter 10 Solutions

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Tofind: the equation and if the equation is parabolic give its vertex, focus and directrix; if its ellipse gives its center, vertices and foci; if it’s is a hyperbola, give its center, vertices, foci and asymptotes.

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Tofind:the equation and if the equation is parabolic give its vertex, focus and directrix; if its ellipse gives its center, vertices and foci; if it’s is a hyperbola, give its center, vertices, foci and asymptotes.

Answer to Problem 3RE

Explanation of Solution

Chapter 10 Solutions

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