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this is not a homework question and I consider this problem quite difficult and confusing. I tried hard to solve it for 2 days, sure I found solutions, but they are not the same that the one provided by the book. I even know the problem by heart, I don't even need to check the book to write it down. The problem is

Transform this parametric equation in rectangular form: $x = a\frac{2t}{1+t^2}$ $y = a\frac{1-t^2}{1+t^2}$

The solution provided by the book is $x^2 + y^2 = a^2$ So I know that this is a circle where a is the radius. It's a circle that moves around the cartesian plane, diagonally, describing a negative slope. I also know that t is a point in the circonference and finally x and y express the position of the circle in the plane.

But the problem is that I found no way to eliminate the parameter. I even tried with a radius a=1 and this doesn't lead me to the desired solution. I always get the parameter into the equation and never manage to eliminate it. I tried similar problems, looked for solutions with online calculators. I even got

$|t| = \frac{\sqrt{(-x+2at)x}}{x}$ or this $t=\frac{2-\sqrt{4-4x^2}}{2x}$ for a=1

But these equations lead to nothing. I tried partial fractions, polar coordinates, algebraic transformations, nothing worked.

I know that it's ok if we, as a students, can't solve some problems, I also know that if I can't solve a problem in 2 days, I would better to drop it and go to the next. But the book solution makes sense and it would be instructive to see how to reach it.

Arturo Magidin
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Casas Particulares Cuba
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  • Mathoverflow is for research-level questions. This question does not belong at the site. You may want to try math.stackexchange, which welcomes questions at all levels. – Arturo Magidin Dec 02 '22 at 18:54
  • These trigonometric identities may help you: https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Tangent_half-angle_substitution – Thomas Dec 02 '22 at 19:35

2 Answers2

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This should be a very easy problem. Try to calculate $x^{2}+y^{2}$ and simplify. See what you get in terms of a.

William Wei
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I remember that

$$\cos^2(x)+\sin^2(x)=1 $$

this identity it is important to have $x^2 + y^2 = a^2$ that is the solution provided by the book. You start from

$$x=a \frac{2t}{1+t^2}, \quad y=a\frac{1-t^2}{1+t^2}\tag 1$$

Incidentally, if you remember the parametric equations in sine and cosine or the so-called Weierstrass substitution, we know

$$\sin(x)=\frac{2t}{1+t^2},\qquad \cos(x)=\frac{1-t^2}{1+t^2}$$

hence the $(1)$ can be written as ($a>0$): $$x=a \sin(x)\iff \sin(x)=\frac xa, \qquad y=a\cos(x)\iff \cos(x)=\frac ya\tag 2$$

which substituted in the fundamental relationship $\cos^2(x)+\sin^2(x)=1$ will have

$$\left(\frac xa\right)^2+\left(\frac ya\right)^2=1 \iff x^2+y^2=a^2\,\,\text{circle di radius a}.$$

This may be a particular way (in this case) to eliminate the parameter $t$ immediately.

Sebastiano
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  • Excellent answer. I didn´t know about the Weierstrass substitution and the problem becomes very easy if you know it. Book authors have a knowledge and assume that everybody have the same knowledge. – Casas Particulares Cuba Dec 03 '22 at 21:34