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Q. 40

Expert-verified
Calculus
Found in: Page 725
Calculus

Calculus

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 (0,±4), directrices y=±1

The equation is y24-x212=1.

See the step by step solution

Step by Step Solution

Step 1. Given information.

We are given,

foci (0,±4), directrices y=±1

Step 2. Value of the variables.

Now, as given,

The focus points are (0,4),(0,-4). Center =0+02,4-42 since mid point =x1+x22,y1+y22 Center =(0,0) Given directries are y=±1. That means be=1b=e Then be=4b·b=4[ since e=b]b2=4c=(0-4)2+(0-0)2 since D=x2-x12+y2-y12 That is the distance from (0,4)(0,0)c=4 For a hyperbola, a2+b2=c2a2+4=42a2=12

Step 3. Substitution.

Now, substitute the obtained values,

(y-k)2b2-(x-h)2a2=1 where (h,k) is the center. (y-0)24-(x-0)212=1 since a2=12,b2=4y24-x212=1

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