
The symmetry in curve ${{x}^{3}}+{{y}^{3}}=3axy$ along
A. X-axis
B. Y-axis
C. Line $y=x$
D. Opposite quadrants
Answer
162.9k+ views
Hint: In this question, we need to find the symmetry of the curve along the given options. By using the symmetric rules of a curve, we are able to find the required condition.
Formula used: The rules of symmetry are:
The variable $y$ in a curve has even power, then it is symmetric about the x-axis.
The variable $x$ in a curve has even power, then it is symmetric about the y-axis.
When we put $x=-x$ and $y=-y$, the curve remains the same, then it is symmetric in opposite quadrants.
The variables $x$ and $y$ are interchanged in a curve, but the curve remains the same. So, we can say that the curve is in symmetry along the line $y=x$.
Complete step by step solution: The given curve is ${{x}^{3}}+{{y}^{3}}=3axy$
Since the curve has odd powers of $x$ and $y$ variables, so the curve is not in symmetry along X-axis and Y-axis.
On substituting $x=-x$ in the curve, we get
$\begin{align}
& {{x}^{3}}+{{y}^{3}}=3axy \\
& \Rightarrow {{(-x)}^{3}}+{{y}^{3}}=3a(-x)y \\
& \Rightarrow -{{x}^{3}}+{{y}^{3}}=-3axy \\
\end{align}$
On substituting $y=-y$ in the curve, we get
\[\begin{align}
& {{x}^{3}}+{{y}^{3}}=3axy \\
& \Rightarrow {{x}^{3}}+{{(-y)}^{3}}=3ax(-y) \\
& \Rightarrow {{x}^{3}}-{{y}^{3}}=-3axy \\
\end{align}\]
Since the curve is changing, it is not in symmetry along the opposite quadrants.
Then, on interchanging$y=x$ or $x=y$ in the curve, we get
\[\begin{align}
& {{x}^{3}}+{{y}^{3}}=3axy \\
& \Rightarrow {{(y)}^{3}}+{{(x)}^{3}}=3a(y)(x) \\
& \Rightarrow {{x}^{3}}+{{y}^{3}}=3axy \\
\end{align}\]
Since the equation of the curve remained the same, it is symmetric along the line $y=x$.
Thus, Option (C) is correct.
Note: Here we need to check all the symmetry conditions for the curve in order to get the actual one. The rules of symmetry are predefined and by using them we can able to find the required condition.
Formula used: The rules of symmetry are:
The variable $y$ in a curve has even power, then it is symmetric about the x-axis.
The variable $x$ in a curve has even power, then it is symmetric about the y-axis.
When we put $x=-x$ and $y=-y$, the curve remains the same, then it is symmetric in opposite quadrants.
The variables $x$ and $y$ are interchanged in a curve, but the curve remains the same. So, we can say that the curve is in symmetry along the line $y=x$.
Complete step by step solution: The given curve is ${{x}^{3}}+{{y}^{3}}=3axy$
Since the curve has odd powers of $x$ and $y$ variables, so the curve is not in symmetry along X-axis and Y-axis.
On substituting $x=-x$ in the curve, we get
$\begin{align}
& {{x}^{3}}+{{y}^{3}}=3axy \\
& \Rightarrow {{(-x)}^{3}}+{{y}^{3}}=3a(-x)y \\
& \Rightarrow -{{x}^{3}}+{{y}^{3}}=-3axy \\
\end{align}$
On substituting $y=-y$ in the curve, we get
\[\begin{align}
& {{x}^{3}}+{{y}^{3}}=3axy \\
& \Rightarrow {{x}^{3}}+{{(-y)}^{3}}=3ax(-y) \\
& \Rightarrow {{x}^{3}}-{{y}^{3}}=-3axy \\
\end{align}\]
Since the curve is changing, it is not in symmetry along the opposite quadrants.
Then, on interchanging$y=x$ or $x=y$ in the curve, we get
\[\begin{align}
& {{x}^{3}}+{{y}^{3}}=3axy \\
& \Rightarrow {{(y)}^{3}}+{{(x)}^{3}}=3a(y)(x) \\
& \Rightarrow {{x}^{3}}+{{y}^{3}}=3axy \\
\end{align}\]
Since the equation of the curve remained the same, it is symmetric along the line $y=x$.
Thus, Option (C) is correct.
Note: Here we need to check all the symmetry conditions for the curve in order to get the actual one. The rules of symmetry are predefined and by using them we can able to find the required condition.
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