randRange(2, 9) randRange(2, 9) A * A B * B randRange( -9, 9 ) randRange( -9, 9 ) rand(2) "+" WHICH_NEG === 1 ? "" : "-" H === 0 ? "x^2" : expr(["^", ["+", "x", -H], 2]) K === 0 ? "y^2" : expr(["^", ["+", "y", -K], 2]) H === 0 ? "\\dfrac{x^2}{" + A2 + "}" : "\\dfrac {" + expr(["^", ["+", "x", -H], 2]) + "}{" + A2 +"}" K === 0 ? "\\dfrac{y^2}{" + B2 + "}" : "\\dfrac {" + expr(["^", ["+", "y", -K], 2]) + "}{" + B2 +"}"

The equation of hyperbola `H` is `WHICH_NEG === 1 ? expr(["-", Y2T, X2T]) : expr(["-", X2T, Y2T]) = 1`.

What are the asymptotes?

`y = \pm` B/A `(x +` -H`) +` K

We want to rewrite the equation in terms of `y`, so start off by moving the `y` terms to one side:

`Y2T = Y_MINUS 1 X_MINUS X2T`

Multiply both sides of the equation by `B2`.

`Y = {Y_MINUS B2 X_MINUS \dfrac{X \cdot B2}{A2}}`

Take the square root of both sides.

`\sqrt{Y} = \sqrt{Y_MINUS B2 X_MINUS \dfrac{X \cdot B2}{A2}}`

`plus("y", -K) = \pm \sqrt{Y_MINUS B2 X_MINUS \dfrac{X \cdot B2}{A2}}`

As `x` approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it.

`plus("y", -K) \approx \pm \sqrt{\dfrac{X \cdot B2}{A2}}`

`plus("y", -K) \approx \pm \left(\dfrac{B \cdot (plus("x", -H))}{A}\right)`

Add `K` to both sides and rewrite Subtract `-K` from both sides and rewrite Rewrite as an equality in terms of `y` to get the equation of the asymptotes:

`y = \pm fractionReduce(B, A)(plus( "x", -H ))K >= 0 ? "+" : "" K`