SOLUTION: Find an equation for the hyperbola with center (-5,3) focus (15,3) and vertex (7,3)

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Question 449574: Find an equation for the hyperbola with center (-5,3) focus (15,3) and vertex (7,3)
Found 2 solutions by MathLover1, FrankM:
Answer by MathLover1(20850) (Show Source):
You can put this solution on YOUR website!

Find an equation for the hyperbola with
center (-5,3) ......... and
focus (15,3) and
vertex (7,3)
The center, focus, and vertex all lie on the horizontal line y = 3 and they're side by side on a line paralleling the x-axis.
So, the branches must be side by side, and the part of the equation
must be added. The will go with the part of the equation, and
the part will be subtracted. The vertex is 12 units from the center, so
; the focus is 8 units from the center, so . Then

gives me .
I don't need to bother with the value of itself, since they only asked
me for the equation, which is:

Answer by FrankM(1040) (Show Source):
You can put this solution on YOUR website!
.
Respectfully, the current answer is incorrect.
center (-5,3) focus (15,3) and vertex (7,3)
Step 1 - Plot these 3 points
Step 2 - Note that c-v distance is 12
Step 3 - Note that c-f distance is 20
Step 4 - create the triangle shown, with vertices of center, vertex, and c-f distance (I rotated that c-f (not drawn) line to be above the vertex)
Step 5 -
Step 6 - and b solves to 16 (I use the traditional a and b are legs and c is hypotenuse. Don't worry if book or teacher tries to label differently. Pythagorus still rules.)
We are done -

Extra credit - show the 2 asymptotes -

and

I use the form above (slope-intercept) as it's relatively easy to get these number to plug in. You see the 3 and the -5 represent the center point. The 12 and 16 are just the a and b numbers, and you only need to be careful to choose the order correctly for rise/run. Then the negative of that slope gives you the other asymptote.