Find an equation of the hyperbola that satisfies the given conditions . Foci (0, +4) one vertex (0,-2) 5. Center at (-2,-4); a vertex at (-2,-7); a focus at (-2,-9) 6. Ends of conjugate axis at (3,5) and (3,-1); eccentricity

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Number 6
Find an equation of the hyperbola that satisfies the given conditions
. Foci (0, +4) one vertex (0,-2)
5. Center at (-2,-4); a vertex at (-2,-7); a focus at (-2,-9)
6. Ends of conjugate axis at (3,5) and (3,-1); eccentricity
Transcribed Image Text:Find an equation of the hyperbola that satisfies the given conditions . Foci (0, +4) one vertex (0,-2) 5. Center at (-2,-4); a vertex at (-2,-7); a focus at (-2,-9) 6. Ends of conjugate axis at (3,5) and (3,-1); eccentricity
Expert Solution
Step 1

Given that, 

Ends of the conjugate axis is at 3,5 and 3,-1

Here the centre of the hyperbola is in the midpoint of 3,5 and 3,-1

Therefore the mid point of two points x1,y1 and x2,y2 is given by

x1+y12, x2+y22

Therefore the mid point of 3,5 and 3,-1 is 3,2

So, the centre of the hyperbola is 3,2.

Therefore the origin is shifted to 3,2.

Let X,Y be the coordinates of the point x,y with respect to the new axis.

Therefore, x=3+X and y=2+Y

Step 2

Therefore ends of the conjugate axis is 

3-3,5-2=0,3 and 3-3,-1-2=0,-3

Therefore b=3

Since the conjugate axis is in y-axis, the foci is in x- axis.

Since the eccentricity, e=ca is given by 54

Then c=5 and a=4

Therefore the hyperbola is 

X216-Y225=1

 

 

 

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