# Hyperbolas on an Imaginary Graph

My first question is what this type of graph (of $x-y-i$) is called since I was unable to find any information about any such graph.

Now for the real question, I used the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and drew the graph on the $xy$ plane and then added an imaginary axis and found out the value $y = bi$. so I plotted it and since it was symmetric I decided to do it for $-b$; then I saw that the values fell sharply so now I think this graph will be an Ellipse. Am I correct?

Is there any research about the xy plane with imaginary axis?

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## 1 Answer

First of all, I would not call it the "imaginary axis" - the name needs to be clearer as to what the value on that axis represents. If I understand your question correctly, it is the imaginary $y$-axis; in other words, you're allowing $y$ to be a complex number $y=y_0+iy_1$, while you are still requiring $x$ to be real.

If I understand your question correctly, you are wondering if your plot of the solutions to $$\frac{x^2}{a^2}-\frac{(y_0+iy_1)^2}{b^2}=1$$ is correct. Well, I'd advise thinking about it like this: $$\frac{x^2}{a^2}-\frac{(y_0+iy_1)^2}{b^2}=\bigg(\frac{x^2}{a^2}-\frac{y_0^2}{b^2}+\frac{y_1^2}{b^2}\bigg)-i\bigg(\frac{2y_0y_1}{b^2}\bigg)=1$$ The only way this is possible is if $y_0=0$ or $y_1=0$; otherwise, the imaginary part of the left side is non-zero, while the imaginary part of the right size is zero.

Thus, you're looking for the solutions to $$\frac{x^2}{a^2}-\frac{y_0^2}{b^2}+\frac{y_1^2}{b^2}=1$$ where either $y_0=0$ or $y_1=0$. The above equation defines a hyperboloid of one sheet, and so you're looking for the intersection of that hyperboloid with the $xy_1$-plane (where $y_0=0$) and $xy_0$-plane (where $y_1=0$).

In conclusion: Your plot seems to be of the right form, though it is not centered correctly (the center of the whole thing should be at the origin).

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