Sketching the curve $\frac{y^2}{a^2}+1=\frac{a}{x}$

Question

Firstly, if don't understand the domain of the function as the constant $a$ is not specified to be positive or negative, (correct me if the sign doesn't make a difference). I tried finding the symmetry of the curve and it comes out to be symmetric about x-axis. (Since $(x,y)$ satifies the equation iff $(x,-y)$ does. Also the derivative turns out to be negative (if we assume only the positive root). (Should we not consider the negative sign of the root). i.e. $y^2=a$, then $y=\pm\sqrt a$. Any help would be appreciated. Thanks.

Answer 1

It may be easier to flip the question around and plot $x$ as a function to $y$.

$$\begin{align} \frac {y^2}{a^2}+1&=\frac ax\\ \frac xa&=\frac{a^2}{y^2+a^2}\\ x&=\frac{a^3}{y^2+a^2}\end{align}$$

Assume $x$ and $y$ are the vertical and horizontal axes respectively.

First plot $x=y^2+a^2$, an upward opening parabola with $x$-intercept of $a^2$.

Take the reciprocal and multiply by $a^3$ to plot $x=\frac{a^3}{y^2+a^2}$. The $x$-intercept is $a$, and the asymptotes are $x=0$ as $y\to \pm\infty$.

Flip the axes around to give the required graph.