If you don't put bounds, I think it's not possible. The graphic of your equation is a cross with four infinite branches and with one paramater, you can describe complicated curves but only with two infinite extensions.

If your variables are bounded, maybe it's possible.

So basically, I wanted to try to make a parametric equation of this equation to infinity and I did some research to transfer an implicit to parametric equation. Still, the case of x^2 + y^2 = 0 is quite complex as you actually said. Because from what I see, when I switch from implicit to parametric, I find some weird things. I actually think you're right, that's impossible.

When you say "If your variables are bounded, maybe it's possible", what do you mean?

1) |cos(t)| / cos(t) = +/- 1 for all values of t ( except when t = multiple of pi/2) so y reduces to +/- sin(t) and x = sin(t) except at t = mult pi/2 when you get a jump discontinuity. But, as t--> pi/2, y --> +/- n.

2) Restricting t to 0<= t <= 2pi only gives you a portion of the 2 lines where x and y are trapped in the square between +/- n. And it doesn't change anything if you let t range over all the real numbers. You are still trapped in the square where x and y are stuck between +/- n.

3) So x = n.sin(t) and y = +/- n.sin(t) except when t = mult of pi/2 where y does not have a value. You might as well use x = n and y = +/-n where you get all of the infinite pairs of solutions and no discontinuities.

x^2 - y^2 = 0 describes 2 straight lines: y = x and y = -x

So let x = t and then y = t or y = -t where t is any real number.

This is sometimes referred to an a "degenerate hyperbola".

For the rectangular hyperbola, x^2 - y^2 = a^2, the asymptotes are the 2 lines y = x and y = -x. As a tends to 0, then the branches of the hyperbola itself actually approach the the 2 lines y = x and and y = -x and when a actually becomes 0 then the hyperbola turns into the 2 lines - hence the term "degenerate"

As a tends to infinity nothing really special happens. The hyperbola still exists has as the same 2 asymptotes, y = x and y = -x . This prevents the hyperbola approaching the 2 vertical lines x = a and x = -a. It just takes larger and larger of values of x to get arbitrarily close to to the asymptotes.

The case where x^2 + y^2 = 0 is the simplest case of all. It describes a single point (0,0)

For the circle x^2 + y^2 = a^2, the radius is a and as a tends to 0 the circle shrinks and approaches the center (0,0) and when a actually takes on the value of zero we get just the point (0,0). Hence a single point is sometimes called a "degenerate circle". As a tends to infinity we just get a larger and larger circle.

The curvature of the circle x^2 + y^2 = a^2 is 1/a so I suppose we can argue the as a tends to infinity the curvature tends to 0 - which is the curvature of a straight line but there is no single equation which can represent such a line.

both of these pairs of equations only pick up the line y=x (i.e. each pair plots the same line) and not the line y=-x and therefore you lose half of the solutions.

The only way to pick up the other half is x = t ; y = -t.

So from above: y=t, x=t or y=-t, x=-t :

You have to change drop one of the - signs to get : y=t, x=t or y=t, x=-t.

If you don't put bounds, I think it's not possible. The graphic of your equation is a cross with four infinite branches and with one paramater, you can describe complicated curves but only with two infinite extensions.

If your variables are bounded, maybe it's possible.

So basically, I wanted to try to make a parametric equation of this equation to infinity and I did some research to transfer an implicit to parametric equation. Still, the case of x^2 + y^2 = 0 is quite complex as you actually said. Because from what I see, when I switch from implicit to parametric, I find some weird things. I actually think you're right, that's impossible.

When you say "If your variables are bounded, maybe it's possible", what do you mean?

Hello ! Sorry to be so late.

If x (and y) are bounded between -n and n (open interval), you may try :

x=n.sin(t) and y=n.tan(t).∣cos(t)∣ for 0 ≤ t ≤ 2𝜋 but excluding 𝜋/4 and 3𝜋/4

I think it's ok : check, please!

Not trying to be a PITA, but:

1) |cos(t)| / cos(t) = +/- 1 for all values of t ( except when t = multiple of pi/2) so y reduces to +/- sin(t) and x = sin(t) except at t = mult pi/2 when you get a jump discontinuity. But, as t--> pi/2, y --> +/- n.

2) Restricting t to 0<= t <= 2pi only gives you a portion of the 2 lines where x and y are trapped in the square between +/- n. And it doesn't change anything if you let t range over all the real numbers. You are still trapped in the square where x and y are stuck between +/- n.

3) So x = n.sin(t) and y = +/- n.sin(t) except when t = mult of pi/2 where y does not have a value. You might as well use x = n and y = +/-n where you get all of the infinite pairs of solutions and no discontinuities.

x^2 - y^2 = 0 describes 2 straight lines: y = x and y = -x

So let x = t and then y = t or y = -t where t is any real number.

This is sometimes referred to an a "degenerate hyperbola".

For the rectangular hyperbola, x^2 - y^2 = a^2, the asymptotes are the 2 lines y = x and y = -x. As a tends to 0, then the branches of the hyperbola itself actually approach the the 2 lines y = x and and y = -x and when a actually becomes 0 then the hyperbola turns into the 2 lines - hence the term "degenerate"

As a tends to infinity nothing really special happens. The hyperbola still exists has as the same 2 asymptotes, y = x and y = -x . This prevents the hyperbola approaching the 2 vertical lines x = a and x = -a. It just takes larger and larger of values of x to get arbitrarily close to to the asymptotes.

The case where x^2 + y^2 = 0 is the simplest case of all. It describes a single point (0,0)

For the circle x^2 + y^2 = a^2, the radius is a and as a tends to 0 the circle shrinks and approaches the center (0,0) and when a actually takes on the value of zero we get just the point (0,0). Hence a single point is sometimes called a "degenerate circle". As a tends to infinity we just get a larger and larger circle.

The curvature of the circle x^2 + y^2 = a^2 is 1/a so I suppose we can argue the as a tends to infinity the curvature tends to 0 - which is the curvature of a straight line but there is no single equation which can represent such a line.

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

both of these pairs of equations only pick up the line y=x (i.e. each pair plots the same line) and not the line y=-x and therefore you lose half of the solutions.

The only way to pick up the other half is x = t ; y = -t.

So from above:

y=t, x=t or y=-t, x=-t :You have to change drop one of the - signs to get :

y=t, x=t or y=t, x=-t.Otherwise you lose the line

y = -x.