# Determining the Line of Symmetry of a Reciprocal Equation.

Problem: How do you determine the line of symmetry of a reciprocal equation?

Solution:

For example, I'll graph the reciprocal function Y=1/(x+2) ^Just a quick sketch

And the equation of the line of symmetry is simply -(x+2), which can be seen here: ^Also a quick sketch

By adding a negative in front of any reciprocal equation, you have the line of symmetry. That much I figured out.

The Real Problem: Why does this happen? I mean, how can I explain it in plain english?

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
First, there is no such thing as "the" line of symmetry. Any hyperbola has two lines of symmetry. In the case you give, y= 1/(x-2), (Your graph is NOT of y= 1/(x+2). That would have a vertical asymptote at x= -2.), the center of the hyperbola is at (2, 0). The line you draw, through (2,0) and perpendicular to the line containing the vertices, so with slope -1, y= -(x-2)= 2- x, is one axis of symmetry. The line through (2,0) and passing through the vertices, y= x- 2, is also an axis of symmetry.

I am not at all sure what you mean by "By adding a negative in front of any reciprocal equation, you have the line of symmetry." Do you mean "if y= 1/(x-a) is a hyperbola, then y= -1(x-a)= a- x is a line of symmetry"? If that is what you mean, then, yes, that is true. For that particular kind of hyperbola, the two lines of symmetry have slopes 1 and -1 and pass through the center of the hyperbola, (a, 0). Their equations are, therefore, y= 1(x-a)= x-a and y=-1(x-a)= a- x.

More generally, any hyperbola has two lines of symmetry: the line through the two vertices and the line through the center (half way between the vertices) perpendicular to that line.