SOLUTION: What's the equation of the hyperbola with a focus at (-3-3*sqrt13,1), asymptotes intersecting at (-3,1) and one asymptote passing through the point (1,7)?

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Question 886273: What's the equation of the hyperbola with a focus at (-3-3*sqrt13,1), asymptotes intersecting at (-3,1) and one asymptote passing through the point (1,7)?
Found 2 solutions by Edwin McCravy, lwsshak3:
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!



So we know that the hyperbola has the equation:

  

The asymptotes intersect at the center, so the center is 

(h,k) = (-3,1).

  

We only need to know a and b.

The slope of the asymptotes are 

So we find the slope of the asymptote through (-3,1) and (1,7).

We use the slope formula:

m = 

m =  =   =  = 

Therefore  



All hyperbolas have the Pythagorean relationship 

We can find the value of " c " because it is the distance from
the focus to the center, so we find the distance 

between the focus  and the
center(-3,1)

using the distance formula:

 











So , and since 



We solve the system of equations:



Solve the second one for b:




Substitute in the first equation:









Substitute in 



Therefore the equation of the hyperbola is:

  

or

  

Edwin
Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
What's the equation of the hyperbola with a focus at (-3-3*sqrt13,1), asymptotes intersecting at (-3,1) and one asymptote passing through the point (1,7)?
***
center: (-3,1) (coordinates of asymptotes intersecting point)
Given hyperbola has a horizontal transverse axis.(center and focus have the same y-coordinate(1)
Standard form of equation for a hyperbola with horizontal transverse axis:
, (h,k)=coordinates of center
..
asymptotes are straight lines that take the form: y=mx+b, m=slope, b=y-intercept
For given hyperbola:
Using given points, (-3,1) and (1,7) on one of the asymptotes:
slope=∆y/∆x=(7-1)/(1-(-3))=6/4=3/2
slope of other asymptote=-3/2
..
slope = b/a=3/2
b=3a/2
c^2=a^2+b^2
c^2=a^2+9a^2/4
c^2=13a^2/4
..
focus=(-3-3√13,1)
c=-3-3√13+3≈10.82
c^2=117
..
117=13a^2/4
13a^2=4*117=468
a^2=468/13=36
b^2=9a^2/4=81
..
Equation of given hyperbola:

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