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Q. 39

Found in: Page 772


Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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Short Answer

Use Cartesian coordinates to express the equations for the hyperbolas determined by the conditions specified in Exercises 38–43.

foci (±6,0), directrices x=±1

The equation is x26-y230=1.

See the step by step solution

Step by Step Solution

Step 1. Given information.

The given values are,

foci (±6,0), directrices x=±1

Step 2. Value of variables.


The focus points are (6,0),(-6,0). Center =6-62,0+02 since mid point =x1+x22,y1+y22 Center =(0,0) Given driectries are x=±1. a=e Then ae=6a2=6c=(0-6)2+(0-0)2sinceD=x2-x12+y2-y12c=6 For a hyperbola, a2+b2=c26+b2=62b2=30

Step 3. Substitution.

Now, substitute the obtained values in,

(x-h)2a2-(y-k)2b2=1 where (h,k) is the center. (x-0)26-(y-0)230=1 since a2=6,b2=30x26-y230=1

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