Precalculus

9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Question

Chapter 10, Problem 19RE

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 19RE

Thecenter= $(h,k)=(1,-1)$ , vertices is $(h±a,k)=(h,k±c)=(1,-1±3)=(1,2) and (1,-4)$ and Foci= $(h,k±c)=(1,-1±5)$ .

Explanation of Solution

Given:

$9x2+4y2-18x+8y=23$ .

Concept used:

Formula for ellipse:

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

The point $(h,k)$ is center of the ellipse.

The coordinates of the vertices are $(±a,0)$

The coordinates of the co-vertices are $(0,±b)$

The coordinates of the foci are $(±c,0)$ where $c2=a2-b2$ .

The length of major and minor axis’s: $2a and 2b$ respectively.

Where $a>b$ .

Calculation:

$9x2+4y2-18x+8y=23$ .

$9x2-18x+9+4y2+8y+4=23+9+4$ .

$9(x2-2x+1)+4(y2+2y+1)=36.$

$9(x-1)236+4(y+1)236=1.$

$(x-1)24+(y+1)29=1.$

Here the equation is in the form of $(x-h)2b2+(y-k)2a2=1$ where $a=3,b=2,h=1,k=-1 and c=a2-b2=5.$ ,it’s an equation of ellipse.

The center= $(h,k)=(1,-1)$ The vertices is $(h±a,k)=(h,k±c)=(1,-1±3)=(1,2) and (1,-4).$

Foci= $(h,k±c)=(1,-1±5)$ .

Hence, center= $(h,k)=(1,-1)$ , vertices is $(h±a,k)=(h,k±c)=(1,-1±3)=(1,2) and (1,-4)$ and Foci= $(h,k±c)=(1,-1±5)$ .

Chapter 10, Problem 18RE

Chapter 10, Problem 20RE

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.