How do you determine whether the graph of y^2=(4x^2)/9-4 is symmetric with respect to the x axis, y axis, the line y=x or y=-x, or none of these?

Dec 24, 2016

The graph is symmetrical, with respect to the axes. There is no symmetry, with respect to the bisectors $y = \pm x$. See the illustrative Socratic graph of this hyperbola.

Explanation:

graph{x^2/8-y^2.4-1=0 [-10, 10, -5, 5]}

The equation is

$f \left(x , y\right) = \frac{4 {x}^{2}}{9} - {y}^{2} - 4 = 0$.

Here, $f \left(\pm x , \pm y\right) = f \left(x , y\right)$.

So, if (x, y) is a point on the graph, then (x, -y), (-x, y) and (-x, -y) lie on

the graph. And so, the graph is symmetrical about both the axes.

$y = \pm x$ become the new axes upon rotation of the axes about the

origin, through ${45}^{o}$. The ad hoc transformations are

$x = \frac{X - Y}{\sqrt{2}} \mathmr{and} y = \frac{X + Y}{\sqrt{2}}$.

Referred to the new X and Y axes, the equation f(x, y) = 0

becomes

$g \left(X , Y\right) = \frac{4}{9} {\left(X - Y\right)}^{2} / 2 - {\left(X + Y\right)}^{2} / 2 - 4 = 0$.

Now, only $g \left(- X , - Y\right) = g \left(X , Y\right)$, revealing, as expected, polar

symmetry about the ( same ) origin.

There is no symmetry about the new axes.

So, there is no symmetry about $y = \pm x$.