Chapter 3.1, Problem 27E

To determine

To find: line(s) of symmetry for the graph of given equation.

Answer to Problem 27E

Symmetric with x-axis, y-axis, $y=x$ and $y=-x$

Explanation of Solution

Given information:

  $x^2=\frac{1}{y^2}$

Calculation:

Put $\left(a,b\right)$ into the equation $\left(xy=1\right)$ and solve as follows:

  $\begin{array}{l}{x}^2=\frac{1}{y^2}\\ a^2=\frac{1}{b^2}\end{array}$

If the solution of equation is same as that of any other case, then their graphs are symmetric.

Let’s check whether the given cases are symmetric with the given equation or not by substituting replacing $\left(x,y\right)$with different coordinates in the equation and simplify as shown below.

Check symmetry with	$\left(x,y\right)$	$x^2=\frac{1}{y^2}$	Symmetric with graph equation (Yes/No)
x-axis	$\left(a,-b\right)$	$\begin{array}{c}{a}^2=\frac{1}{{\left(-b\right)}^2}\\ a^2=\frac{1}{b^2}\end{array}$	Yes
y-axis	$\left(-a,b\right)$	$\begin{array}{c}{\left(-a\right)}^2=\frac{1}{b^2}\\ a^2=\frac{1}{b^2}\end{array}$	Yes
$y=x$	$\left(b,a\right)$	$\begin{array}{c}{b}^2=\frac{1}{a^2}\\ \text{or, }{a}^2=\frac{1}{b^2}\end{array}$	Yes
$y=-x$	$\left(-b,-a\right)$	$\begin{array}{c}{\left(-b\right)}^2=\frac{1}{{\left(-a\right)}^2}\\ b^2=\frac{1}{a^2}\\ \text{or, }{a}^2=\frac{1}{b^2}\end{array}$	Yes

Thus the graph of equation is symmetric with x-axis, y-axis, $y=x$ and $y=-x$ .