Q 11 - Precalculus Enhanced with Graphing Utilities

Question

Test each equation for symmetry with respect to the x-axis, the y-axis, and the origin.  

$x^2+4y^2=16$.

Step-by-Step Solution

Answer

The given equation is symmetric about x-axis, y-axis and origin.

Step 1. Given Information

We have given the following equation :- 

$x^2+4y^2=16$.

We have to check the symmetry of this equation with respect to x-axis, y-axis and the origin. 

Step 2. To check symmetry about x-axis.

The given equation is :- 

$x^2+4y^2=16$.

We know that a graph is symmetrical about x-axis, if a point $\left(x,y\right)$ lies on graph, then $\left(x,-y\right)$ is also lies on graph.

So to check symmetry about x-axis, change $y$ by $-y$ in  the given equation, then we have :-

$$\begin{gathered} x^2+4{(-y)}^2=16 \\ \Rightarrow x^2+4y^2=16 \end{gathered}$$.

The resulting equation is same as the given equation.

So we can say that the given equation is symmetric about x-axis.

Step 3. To check symmetry about y-axis.

The given equation is :- 

$x^2+4y^2=16$.

We know that a graph is symmetrical about y-axis, if a point $\left(x,y\right)$ lies on graph, then $\left(-x,y\right)$ is also lies on graph.

So to check symmetry about y-axis, change $x$ by $-x$ in the given equation, then we have :-

$$\begin{gathered} {(-x)}^2+4y^2=16 \\ \Rightarrow x^2+4y^2=16 \end{gathered}$$

The resulting equation is same as the given equation.

So we can say that the given equation is symmetric about y-axis.

Step 4. To check symmetry about origin

The given equation is :- 

$x^2+4y^2=16$.

We know that a graph is symmetrical about origin, if a point $\left(x,y\right)$ lies on graph, then $\left(-x,-y\right)$ is also lies on graph.

So to check symmetry about origin, change $x$ by $-x$ and $y$ by $-y$ in the given equation, then we have :-

$$\begin{gathered} {(-x)}^2+4{\left(-y\right)}^2=16 \\ \Rightarrow x^2+4y^2=16 \end{gathered}$$

The resulting equation is same as the given equation.

So we can say that the given equation is symmetric about origin.