Determine whether the graph of the equation is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

[tex]\bf x^2y^2+3xy=1\\\\ -------------------------------\\\\ \stackrel{\stackrel{\textit{test for x-symmetry}}{y=-y}}{x^2(-y)^2+3x(-y)=1}\implies x^2y^2-3xy=1\impliedby \begin{array}{llll} \textit{function differs}\\ \textit{from original}\\ \textit{no dice} \end{array}\\\\ -------------------------------\\\\[/tex]

[tex]\bf \stackrel{\stackrel{\textit{test for y-symmetry}}{x=-x}}{(-x)^2y^2+3(-x)y=1}\implies x^2y^2-3xy=1\impliedby \begin{array}{llll} \textit{function differs}\\ \textit{from original}\\ \textit{no dice} \end{array}\\\\ -------------------------------\\\\ \stackrel{\stackrel{\textit{test for origin-symmetry}}{x=-x~~y=-y}}{(-x)^2(-y)^2+3(-x)(-y)=1}\implies x^2y^2+3xy=1\impliedby \begin{array}{llll} origin\\symmetry \end{array}[/tex]

so, recall, the function has symmetry when the yielded resulting function resembles the original function, after negativizing the variable(s).

Also recall that minus*plus is minus, and minus*minus is plus.

KarpaT KarpaT · Answer 2 · 2022-06-15T18:27:49+02:00

The equation is symmetric about origin and it is graphically shown.

What is a symmetry of a function?

A symmetry of a function is a transformation that leaves the graph unchanged. Symmetric function is a function having several variables which remain unchanged for any type of permutation of the variable.

For the given situation,

The equation is

[tex]x^2y^2 + 3xy = 1 -------- (1)[/tex]

Plot this equation on the graph as shown.

To find the symmetry on x-axis, substitute the points as (x,-y)

⇒ [tex]x^2(-y)^2 + 3x(-y) = 1[/tex]

⇒ [tex]x^2y^2 - 3xy = 1 ------ (2)[/tex]

Equation 1 ≠ Equation 2,

Thus this equation is not symmetric to x-axis.

To find the symmetry on y-axis, substitute the points as (-x,y)

⇒ [tex](-x)^2y^2 + 3(-x)y = 1[/tex]

⇒ [tex]x^2y^2 - 3xy = 1 ------ (3)[/tex]

Equation 1 ≠ Equation 3,

Thus this equation is not symmetric to y-axis.

To find the symmetry on origin, substitute the points as (-x,-y)

⇒ [tex](-x)^2(-y)^2 + 3(-x)(-y) = 1[/tex]

⇒ [tex]x^2y^2 + 3xy = 1 -------- (4)[/tex]

Equation 1 = Equation 4,

Thus this equation is symmetric about origin.

Hence we can conclude that the equation is symmetric about origin.

Learn more about symmetry of a function here

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