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)?

Algebra ->  Quadratic-relations-and-conic-sections -> 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)?      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!



So we know that the hyperbola has the equation:

%28x-h%29%5E2%2Fa%5E2%22%22-%22%22%28y-k%29%5E2%2Fb%5E2 %22%22=%22%22 1

The asymptotes intersect at the center, so the center is 

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

%28x%2B3%29%5E2%2Fa%5E2%22%22-%22%22%28y-1%29%5E2%2Fb%5E2 %22%22=%22%22 1

We only need to know a and b.

The slope of the asymptotes are %22%22+%2B-+b%2Fa

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

We use the slope formula:

m = %28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29

m = %287-1%29%2F%281-%28-3%29%29 =  6%2F%281%2B3%29 = 6%2F4 = 3%2F2

Therefore  

b%2Fa%22%22=%22%223%2F2

All hyperbolas have the Pythagorean relationship c%5E2=a%5E2%2Bb%5E2

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 %28matrix%281%2C3%2C-3-3sqrt%2813%29%2C%22%2C%22%2C1%29%29 and the
center(-3,1)

using the distance formula:

d%22%22=%22%22+sqrt%28+%28x%5B2%5D-x%5B1%5D%29%5E2+%2B+%28y%5B2%5D-y%5B1%5D%29%5E2%29 

c%22%22=%22%22+sqrt%28+%28-3-%28-3-3sqrt%2813%29%29%29%5E2+%2B+%281-1%29%5E2%29

c%22%22=%22%22+sqrt%28+%28-3%2B3%2B3sqrt%2813%29%29%5E2+%2B+%280%29%5E2%29

c%22%22=%22%22+sqrt%28+%283sqrt%2813%29%29%5E2%29+

c%22%22=%22%22sqrt%289%2A13%29

c%22%22=%22%22sqrt%28117%29

So c%5E2=117, and since c%5E2=a%5E2%2Bb%5E2

117=a%5E2%2Bb%5E2

We solve the system of equations:

system%28117=a%5E2%2Bb%5E2%2Cb%2Fa=3%2F2%29

Solve the second one for b:

2b=3a
b=3a%2F2

Substitute in the first equation:

117=a%5E2%2B%283a%2F2%29%5E2
117=a%5E2%2B9a%5E2%2F4
468=4a%5E2%2B9a%5E2
468=13a%5E2
468%2F13=a%5E2
36=a%5E2
6=a

Substitute in b=3a%2F2

b=3%286%29%2F2=9

Therefore the equation of the hyperbola is:

%28x%2B3%29%5E2%2F6%5E2%22%22-%22%22%28y-1%29%5E2%2F9%5E2 %22%22=%22%22 1

or

%28x%2B3%29%5E2%2F36%22%22-%22%22%28y-1%29%5E2%2F81 %22%22=%22%22 1

Edwin

Answer by lwsshak3(11628) About Me  (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:
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1, (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:
%28x%2B3%29%5E2%2F36-%28y-1%29%5E2%2F81=1