Which locus does it represent?

Question

I want to find the following locus $$ y=\frac{k\cdot a\cdot(a\cdot x+j)}{1+a^2\cdot x^2}$$ where $k$ and $a$ are real constants and $x$ is the real variable and $j$ is the imaginary number. How can I determine the locus of $y$ when $x$ vary? I am blocked. Thanks for any help

Answer 1

First of all, we can drop factor $k$ which is responsible for a final enlargment that we can manage at the end.

Then, let us modify notations in order to call things by their usual name:

Consider that we are looking for the locus of

$$\tag{1}z=\underbrace{\frac{a^2 \cdot t}{1+a^2\cdot t^2}}_{x}+\underbrace{\frac{a}{1+a^2\cdot t^2}}_y \cdot j$$

(the former $x$ is now called $t \in \mathbb{R}$).

It is not difficult, by taking a common denominator $2(1+a^2t^2)^2$, to see that

$$\tag{2}x^2+\left(y-\dfrac{a}{2}\right)^2=\dfrac{a^2}{4}$$

which expresses the fact that the locus of $z$ is included into the circle with center $(0,\dfrac{a}{2})$, and radius $\dfrac{a}{2}$. A quick analysis shows that it is the whole circle excepted point $(0,0)$.

Final multiplication by $k$ yields an homothetic circle with center $(0,\dfrac{ka}{2})$, and radius $\dfrac{ka}{2},$ deprived of a certain point.

Edit: it may have some interest to understand that this exercice has to do with the (classical) parameterization of an algebraic curve by using a sweeping line through a point of this curve. Have a look at the picture below: the sweeping line has equation $y=\dfrac{x}{ta}$ ; its 2nd point of intersection with the circle having equation (2) has coordinates :

$$\begin{cases}x & = & \frac{a^2 \cdot t}{1+a^2\cdot t^2}\\ y & = & \frac{a}{1+a^2\cdot t^2}\end{cases}$$

i.e., the expressions of $x$ and $y$ that one has in (1).

This explains in a natural way why point $(0,0)$ is not part of the locus.

Answer 2

Hint: $\;\;y=\cfrac{k\cdot a\cdot(a\cdot x+j)}{1+a^2\cdot x^2}=\cfrac{ka}{ax-j}\,$, so $\,\cfrac{ka}{y}=ax-j\,$, then:

$$ka\left(\cfrac{1}{y}-\cfrac{1}{\overline{y}}\right)=-2j \;\iff\; 0 = y \bar y + \frac{ka}{2j}(\bar y- y)= \left(y+\frac{ka}{2j}\right)\left(\bar y-\frac{ka}{2j}\right)-\frac{k^2a^2}{4}$$