Chapter 1.2, Problem 85AYU

To determine

To find: The intercepts of the graph of the equation ${(x^2 + y^2 - x)}^2 = x^2 + y^2$ and symmetry with respect to $x\text{-axis}$, $y\text{-axis}$ and the origin.

Answer to Problem 85AYU

The $x\text{-intercept}$ are 0 and 2; the $y\text{-intercept}$ are $-1$ and 1

The graph of the equation is symmetric with respect to $x\text{-axis}$

Explanation of Solution

Given:

The equation ${(x^2 + y^2 - x)}^2 = x^2 + y^2$

Calculation:

To find the $x\text{-intercept}$ let $y = 0$ in the equation

${(x^2 + y^2 - x)}^2 = x^2 + y^2$

${(x^2 + 0^2 - x)}^2 = x^2 + {(0)}^2$

${(x^2 - x)}^2 = x^2$

${[x (x - 1)]}^2 = x^2$   Taking $x$ common factor inside the bracket

$x^2{(x- 1)}^2 = x^2$

${(x - 1)}^2 = 1$   Dividing both sides by $x^2$

$x^2 - 2x + 1 = 1$   Expanding the left hand side

$x^2 - 2x + 1 - 1 = 1 - 1$   Subtracting 1 from both sides

$x^2 - 2x = 0$

$x\left(x-2\right)=0$   Taking $x$ common on left hand side

$x = 0$ or $x\mathrm{ - 2} = 0$

$x= 0$ or $x =\text{ 2}$

The $x\text{-intercept}$ are 0 and 2

To find the $y\text{-intercept}$ let $x = 0$ in the equation

${(x^2 + y^2 - x)}^2 = x^2 + y^2$

${(0^2 + y^2 - 0)}^2 = 0^2 + y^2$

${(y^2)}^2 = y^2$

$y^4 = y^2$

$y^2 = 1$   Divide both sides by $y^2$

$y = \pm 1$

The $y\text{-intercepts}$ are $\mathrm{-1}$ and 1

To test the symmetry of the graph with respect to $x\text{-axis}$, replace $y$ by $-y$ in the equation

${(x^2 + {(-y)}^2 - x)}^2 = x^2 + {(-y)}^2$

${(x^2+y^2-x)}^2=x^2+y^2$   which is equivalent to the original equation.

The graph of the equation is symmetric with respect to the $x\text{-axis}$.

To test the symmetry of the graph with respect to $y\text{-axis}$, replace $x$ by $-x$ in the equation

${({(-x)}^2+y^2-(-x))}^2={(-x)}^2+y^2$

${(x^2+y^2+x)}^2=x^2+y^2$   which is not equivalent to the original equation.

The graph of the equation is not symmetric with respect to the $y\text{-axis}$.

To test the symmetry of the graph with respect to the origin, replace $x$ by $x$ and $y$ by $-y$ in the equation

${({(-x)}^2+{(-y)}^2-(-x))}^2={(-x)}^2+{(-y)}^2$

${(x^2+y^2+x)}^2=x^2+y^2$   which is not equivalent to the original equation.

The graph of the equation is not symmetric with respect to the origin.