Algebra - Symmetry

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Section 4.7 : Symmetry

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2. Determine the symmetry of each of the following equation.

$\frac{{{y^2}}}{4} = 1 + \frac{{{x^2}}}{9}$

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Let’s first check for symmetry about the $x$ -axis. To do this we need to replace all the $y$ ’s with – $y$ .

$\frac{{{{\left( { - y} \right)}^2}}}{4} = 1 + \frac{{{x^2}}}{9}\hspace{0.25in} \to \hspace{0.25in}\frac{{{y^2}}}{4} = 1 + \frac{{{x^2}}}{9}$

The resulting equation is identical to the original equation and so is equivalent to the original equation. Therefore, the equation is has symmetry about the $x$ -axis.

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Next, we’ll check for symmetry about the $y$ -axis. To do this we need to replace all the $x$ ’s with – $x$ .

$\frac{{{y^2}}}{4} = 1 + \frac{{{{\left( { - x} \right)}^2}}}{9}\hspace{0.25in} \to \hspace{0.25in}\frac{{{y^2}}}{4} = 1 + \frac{{{x^2}}}{9}$

The resulting equation is identical to the original equation and so is equivalent to the original equation. Therefore, the equation is has symmetry about the $y$ -axis.

Show Step 3

Finally, a check for symmetry about the origin. For this check we need to replace all the $y$ ’s with – $y$ and to replace all the $x$ ’s with – $x$ .

$\frac{{{{\left( { - y} \right)}^2}}}{4} = 1 + \frac{{{{\left( { - x} \right)}^2}}}{9}\hspace{0.25in} \to \hspace{0.25in}\frac{{{y^2}}}{4} = 1 + \frac{{{x^2}}}{9}$

The resulting equation is identical to the original equation and so is equivalent to the original equation. Therefore, the equation is has symmetry about the origin.