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?

1 Answer
Dec 24, 2016

Answer:

The graph is symmetrical, with respect to the axes. There is no symmetry, with respect to the bisectors #y=+-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(x, y)=(4x^2)/9-y^2-4=0#.

Here, #f(+-x, +-y)=f(x, y)#.

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=+-x# become the new axes upon rotation of the axes about the

origin, through #45^o#. The ad hoc transformations are

#x =(X-Y)/sqrt2 and y=(X+Y)/sqrt2#.

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

becomes

#g(X, Y)=4/9(X-Y)^2/2-(X+Y)^2/2-4=0#.

Now, only #g(-X, -Y)=g(X, Y)#, revealing, as expected, polar

symmetry about the ( same ) origin.

There is no symmetry about the new axes.

So, there is no symmetry about #y = +-x#.