NB: very important - √(t^2), not t because, in effect, it divides by -1 when the arg ≥ π

Derivation:

Make x the subject:

x^2-y^2=0

y^2=-x^2

x=±y

Now we create a method to have 2 sets of points i.e. y=x, y=-x:

t is our parameter.

let r = |t|

let θ = arg(t)

Therefore t = r * cis(θ)

let s = t^2 = r^2 * cis(2θ)

let d = t / √(s)

Therefore d = 1 for 0 ≤ θ < π

-1 for π ≤ θ < 2π

This (using d) gives us a split into 2 sections, which are the same size as the original.

So now, we do the following:

let n = r * cis(2θ) = s / |t|

Now n is the same number at double the angle, which allows us to get any complex number twice from the value of t, which is what we need, it also repeats where the value of d switches from 1 to -1, which is what we need.

let x = n

let y = dn

With this setup, we get every number twice, and then flip the sign for every y value so we get every number paired with the number, and the negated version of this number.

Therefore we have parameterised the relation.

Simplify to only use the pronumeral t:

x = n

= s / |t|

= t^2 / |t|

y = dn

= d * s / |t|

= d * t^2 / |t|

= (t / √(s)) * t^2 / |t|

= t^2 * (t / √(t^2)) / |t|

= t^3 / √(t^2) / |t|

= t^3 / (√(t^2) * |t|)

Therefore, it is possible to parameterise this relation over the complex plane with only one value for x and y for each value of t.

If, for example, you set a = 2 then you get x =2t and y = 2t. Let t range over the real numbers you only get one of the 2 straight lines (y =x). You have to include y = +/- at in order to pick up both branches. Same holds true if a is negative. It's still true that x^2 - y^2 = 0 but with x = at and y = at you only pick up half the solutions --- y=x.

x = at ; y = +/-at is the only way to pick up both lines. The "a" is unnecessary here but it doesn't hurt either. It's just simpler to exclude it.

Here's a way to do it over the complex plane:

Parameterised form:x = t^2 / |t|

y = t^3 / (√(t^2) * |t|)

NB: very important - √(t^2), not t because, in effect, it divides by -1 when the arg ≥ π

Derivation:Make x the subject:

x^2-y^2=0

y^2=-x^2

x=±y

Now we create a method to have 2 sets of points i.e. y=x, y=-x:

t is our parameter.

let r = |t|

let θ = arg(t)

Therefore t = r * cis(θ)

let s = t^2 = r^2 * cis(2θ)

let d = t / √(s)

Therefore d = 1 for 0 ≤ θ < π

-1 for π ≤ θ < 2π

This (using d) gives us a split into 2 sections, which are the same size as the original.

So now, we do the following:

let n = r * cis(2θ) = s / |t|

Now n is the same number at double the angle, which allows us to get any complex number twice from the value of t, which is what we need, it also repeats where the value of d switches from 1 to -1, which is what we need.

let x = n

let y = dn

With this setup, we get every number twice, and then flip the sign for every y value so we get every number paired with the number, and the negated version of this number.

Therefore we have parameterised the relation.

Simplify to only use the pronumeral t:

x = n

= s / |t|

= t^2 / |t|

y = dn

= d * s / |t|

= d * t^2 / |t|

= (t / √(s)) * t^2 / |t|

= t^2 * (t / √(t^2)) / |t|

= t^3 / √(t^2) / |t|

= t^3 / (√(t^2) * |t|)

Therefore, it is possible to parameterise this relation over the complex plane

with only one value for x and y for each value of t.Sorry, I missed the addendum:

Also y=-at, x=at, y=at, x=-at, and y=-at, x=-at?y=-at, x=at or y=at,x=at is all that is needed.

The comma is an implied "and" and the "or" is necessary.

*x=-at in above

Also y=-at, x=at, y=at, x=-at, and y=-at, x=-

y=at, x=at, a not equal to zero. It is such a easy question. Why create complexity.

y=t, x=t or y=-t, x=-t.