Instead of doing your problem for you, I'm doing one which is EXACTLY
STEP-BY-STEP like yours. All you have to do is use it as a model and
follow it step-by-step. Here's the problem that I'll solve for you:
Find the equation of the hyperbola (in standard and general forms) which
satisfies the conditions given.
Foci at (4-√41,5) and (4+√41,5), vertices at (9,5) and (-1,5)
We plot the center and foci to see whether the hyperbola is like
this ) ( or like this:
The hyperbola is like this ") (",so its equation is of this form:
where (h,k) = the center = (4,5),
and "a" = the distance from the center to the vertex, which is the
distance from (4,5) to (9,5) which is 9-4=5
so we have this much of the
equation:
The distance from the center to either focus is the value c,
and that is the distance from (4,5) to (4+√41,5), which is
(4+√41)-(4) = √41
and the Pythagorean relationship for all hyperbolas is
c2=a2+b2.
<---that's the standard form.
That's the standard form equation. To draw the graph, we make a defining
rectangle which is 2a=10 units wide and 2b=8 units high, with the center (4,5)
as the center of the defining rectangle, and draw its extended diagonals,
which are the asymptotes.
Now we can sketch in the hyperbola:
To get the general form we start with the standard form:
We multiply through by the LCD = (25)(16) = 400
<--that's the general form.
Now go do yours the exact same way, step by step.
You weren't asked to draw the graph, but it helps.
Edwin