How do I identify the symmetry of the graph (3x^2)+(3y^2)=5?

2 Answers
Dec 17, 2017

x= 0

y = mx , m in RR

( m can be any real value )

Explanation:

We can see by dividing by 5:

x^2 + y^2 = 5/3

=> x^2 + y^2 = (sqrt(5/3))^2

We know the general equation of a cirlce:

x^2 + y^2 = R^2 is a cirlce of centre (0,0) of radius R

Hence this is a circle of radius, sqrt(5/3)

A sketch:

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We can imediately see that when flipped in the x and y axis, it stays the same:

y = 0 , x = 0 are the first two lines of symmetry we can see

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We can also see by this sketch, that any line of the equation y = mx is a line of symetry, as when fliped, it stays the same

=> y = mx , m in RR are also lines of symmetry

Dec 17, 2017

It is a circle, so has symmetry group O(2)

Explanation:

Given:

3x^2+3y^2=5

Divide both sides by 3 to find:

x^2+y^2 = 5/3

That is:

(x-0)^2+(y-0)^2 = (sqrt(15)/3)^2

The standard form of the equation of a circle with centre (h, k) and radius r is:

(x-h)^2+(y-k)^2 = r^2

So we can see that the given equation is that of a circle with centre (h, k) = (0, 0) and radius r=sqrt(15)/3

graph{x^2+y^2=5/3 [-4.4, 4.4, -2.2, 2.2]}

This has reflective symmetry about any line through the centre.

It also has rotational symmetry by any angle about the centre. In fact any rotation can be generated by a combination of two reflections.

The symmetry group of these reflections/rotations is called O(2) - the orthogonal group in two dimensions.