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

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

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Answer 1 · 2016-12-24T11:35:31

The graph is symmetrical, with respect to the axes. There is no symmetry, with respect to the bisectors $y = \pm x$ $y=+-x$ . 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

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

Here, $f (\pm x, \pm y) = f (x, y)$ $f(+-x, +-y)=f(x, y)$ .

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.

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

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

$x = \frac{X - Y}{\sqrt{2}} and y = \frac{X + Y}{\sqrt{2}}$ $x =(X-Y)/sqrt2 and y=(X+Y)/sqrt2$ .

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

becomes

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

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

symmetry about the ( same ) origin.

There is no symmetry about the new axes.

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

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

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Explanation:

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