Hyperbola / Rotated Hyperbola Intersection

Question

I am trying to find the point where two hyperbolas intersect, that is, to find a vertex that is common to both hyperbolas.

Also, note that I am only testing for a region of both hyperbolas -- only a quarter of each (either $-b/a$, $b/a$, $-a/b$ or $a/b$).

Here is what I have so far for getting any vertex on a hyperbola: $(x,y)=(a\ cosh(r),b\ sinh(r))$

Now suppose I have two hyperbolas: -one at the origin $(h,k)=(0,0)$ and not rotated($t=0$) -one not at the origin $(h,k)=(3,10)$ and rotated ($t=\pi/4$)

How would you go about and compare the two?

My guess is to start with something like this: $(a_1cosh(r_1),b_1sinh(r_1))=(a_2cosh(r_2),b_2sinh(r_2))$

And then isolate $r_1$ and $r_2$, but then what does it look like when a translation and rotation is involved?

Or maybe I am completely off track.

Answer 1

Let us consider the following two hyperbolas:

The canonical equation of a hyperbola is:

x^2/a^2 + y^2/b^2 = 1

Now, let's consider two hyperbolas:

x^2/a^2 + y^2/b^2 = 1
x^2/c^2 + y^2/d^2 = 1

You determine y^2 based on x^2 from the first equation:

y^2 = b^2(1 - x^2/a^2)

You use this information in the second equation:

x^2/c^2 + b^2(1 - x^2/a^2)/d^2 = 1

From here:

x^2/c^2 + b^2/d^2 - (b^2/(a^2 * d^2))x^2 = 1

Therefore:

(a^2 * d^2 - b^2 * c^2)x^2/(a^2 * c^2 * d^2) = 1 - b^2/d^2

Finally:

x^2 = ((1 - b^2) / d^2) * ((a^2 * c^2 * d^2) / (a^2 * d^2 - b^2 * c^2))

Simplifying it:

x^2 = ((1 - b^2) * a^2 * c^2) / (a^2 * d^2 - b^2 * c^2))

You can now calculate the square root of x and use it for y^2 as well, finally square root y. Do not forget the +- values.