Hi Melody
Here's the method I used, (though later I realised that there was something a whole lot quicker), and btw, there's a typo in the solution I gave earlier, the x term should be negative, -448x, apologies.
Just the framework, I'll leave you to fill in the algebra if you wish.
Start by shifting the intersection of the two lines to the origin,
X = x - 32/7, Y = y - 8/7.
The new lines are Y = 5X/6 and Y = 13X/10.
Assume that the equation of the hyperbola is Y^2 + aX^2 + bXY + c = 0.
(Can divide through by the coefficient of the Y^2 term so that it appears as 1 in the equation).
We need to see what happens to this as X and Y go to infinity.
Divide throughout by X^2,
(Y/X)^2 + b(Y/X) + a + c/X^2 = 0.
As X and Y go to infinity, this becomes (Y/X)^2 + b(Y/X) + a = 0, and that should give us the equations of the two asymptotes.
Using the usual formula for the solution of a quadratic,
(Y/X) = (-b +- sqrt(b^2 - 4a)) / 2.
The plus sign should give us the larger of the two gradients 13/10 and the negative sign the smaller, 5/6.
Solve for a and b. a = 13/12, b = -32/15.
Now substitute into the earlier equations.
That gets you the equation I posted earlier (except for the typo I mentioned).
It occured to me when I'd been through this that I could have arrived at the result simply by multiplying the equations of the two lines together,
(5x - 6y - 16)(13x - 10y - 48) = 0,
though the meaning of the constant term 768 is interesting.
- Bertie
