# A hyperbola as a constant difference of distances

I understand that a hyperbola can be defined as the locus of all points on a plane such that the absolute value of the difference between the distance to the foci is $2a$, the distance between the two vertices.

In the simple case of a horizontal hyperbola centred on the origin, we have the following:

1. $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

2. $c = \sqrt{a^2 + b^2} = a\varepsilon = a\sqrt{1 + \frac{b^2}{a^2}}$

The foci lie at $(\pm c, 0)$.

Now, if I'm not wrong about that, then this should be pretty basic algebra, but I can't see how to get from the above to an equation given a point $(x,y)$ describing the difference in distances to the foci as being $2a$. While I actually do care about the final result, how to get there is more important.

Why do I want to know this? Well, I'd like to attempt trilateration based off differences in distance rather than fixed radii.

-

1
What is "an equation that results in $2a$ given x and y"? – anon Jul 5 '11 at 1:29
@anon I've reworded that bit in a way that might make more sense. – Iskar Jarak Jul 5 '11 at 2:13

## 2 Answers

We will use a little trick to avoid work. We want to have $$\sqrt{(x+c)^2+y^2} -\sqrt{(x-c)^2+y^2}=\pm 2a.\qquad\text{(Equation 1)}$$

Rationalize the numerator, by multiplying "top" and "bottom" by $\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}.$ After the (not very dense) smoke clears, we get $$\frac{4xc}{\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}}=\pm 2a.$$ Flip it over, do some easy algebra. We get $$\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}=\pm \frac{2cx}{a}.\qquad\text{(Equation 2)}$$ From Equations 1 and 2, by adding, we get $$2\sqrt{(x+c)^2+y^2}=\pm 2\left(a+ \frac{xc}{a}\right).$$ Cancel the $2$'s, square. We get $$x^2+2cx+c^2+y^2=a^2+ 2cx+ \frac{c^2x^2}{a^2}.$$ Now it's basically over, the $2cx$ terms cancel. Multiply through by $a^2$, put $c^2=a^2+b^2$, and rearrange.

-

1
+1 much better than mine – Ross Millikan Jul 5 '11 at 3:33
@Ross Millikan: Not really. But "rationalizing the numerator" does have a number of uses, so it is nice to be able to mention it. – André Nicolas Jul 5 '11 at 3:38
Thanks, that was a big help. Totally forgot about that rationalising trick. – Iskar Jarak Jul 5 '11 at 3:51

If we write the equation for what you said, a point $(x,y)$ on the hyperbola, taking $x \gt 0$ for convenience, must have $\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=2a$. Squaring, $(x+c)^2+2y^2+(x-c)^2-2\sqrt{((x-c)^2+y^2)((x+c)^2+y^2)}=4a^2$. Then if you isolate the radical and square again, you should be able to cancel a lot of terms and get to the form you want.

-

This was still helpful although not as detailed as the answer I accepted - I'd upvote if I could. – Iskar Jarak Jul 5 '11 at 3:51
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9509273767471313, "perplexity_flag": "middle"}