Question

How do you convert the parametric equations into a Cartesian equation by eliminating the parameter
$r$: $x = \left( {{r^2}} \right) + r,y = \left( {{r^2}} \right) - r$?

Answer 1

Here, we will first add the given parametric equations and simplify it to get the value of $r$. Then we will multiply the given parametric equations and simplify it further. Then we will substitute the obtained value of $r$ in the equation to eliminate the parameters. Simplifying the equation we will find an equation without the parameters.

Formula Used:
The square of the sum of two numbers is given by an algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

Complete Step by Step Solution:
We are given with the parametric equations $x = \left( {{r^2}} \right) + r$ and $y = \left( {{r^2}} \right) - r$.
Now, we will add the given two parametric equations. Therefore, we get
$x + y = \left( {{r^2}} \right) + r + \left( {{r^2}} \right) - r$
By adding and subtracting the like terms, we get
$\Rightarrow x + y = 2{r^2}$
Dividing both sides by 2, we get
$\Rightarrow {r^2} = \dfrac{1}{2}\left( {x + y} \right)$…………………………………………………………………………………………………..$\left( 1 \right)$
Now, we will multiply the given two parametric equations. Therefore, we get
$xy = \left( {\left( {{r^2}} \right) + r} \right)\left( {\left( {{r^2}} \right) - r} \right)$
We will multiply the expression by using the FOIL method. So, we get
$\Rightarrow xy = {r^2}\left( {\left( {{r^2}} \right) + r} \right) + r\left( {\left( {{r^2}} \right) - r} \right)$
Now, we will multiply each term in the equation, we get
$\Rightarrow xy = {r^4} + {r^3} - {r^3} - {r^2}$
Adding and subtracting the like terms, we get
$\Rightarrow xy = {r^4} - {r^2}$…………………………………………….$\left( 2 \right)$
Now, by substituting the equation $\left( 1 \right)$ in equation $\left( 2 \right)$, we get
$\Rightarrow xy = {\left( {\dfrac{1}{2}\left( {x + y} \right)} \right)^2} - \dfrac{1}{2}\left( {x + y} \right)$
Applying the exponent on the terms, we get
$\Rightarrow xy = \dfrac{1}{4}{\left( {x + y} \right)^2} - \dfrac{1}{2}\left( {x + y} \right)$
By taking L.C.M. to equalize the denominator, we get
$\Rightarrow xy = \dfrac{1}{4}{\left( {x + y} \right)^2} - \dfrac{1}{2} \times \dfrac{2}{2}\left( {x + y} \right)$
$\Rightarrow xy = \dfrac{1}{4}{\left( {x + y} \right)^2} - \dfrac{2}{4}\left( {x + y} \right)$
Subtracting the terms, we get
$\Rightarrow xy = \dfrac{{{{\left( {x + y} \right)}^2} - 2\left( {x + y} \right)}}{4}$
Multiplying both sides by 4, we get
$\Rightarrow 4xy = {\left( {x + y} \right)^2} - 2\left( {x + y} \right)$
The square of the sum of two numbers is given by an algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
Now, by using the algebraic identity and by simplifying, we get
$\Rightarrow 4xy = {x^2} + {y^2} + 2xy - 2x - 2y$
Rewriting the equation, we get
$\Rightarrow {x^2} + {y^2} + 2xy - 2x - 2y - 4xy = 0$
Subtracting the like terms, we get
$\Rightarrow {x^2} + {y^2} - 2x - 2y - 2xy = 0$

Therefore, the Cartesian equation for the given parametric equation by eliminating the Parameters is ${x^2} + {y^2} - 2x - 2y - 2xy = 0$.

Note:
We know that the parametric equation is defined as a function that depends on the Independent variable which is called parameters and the dependent variable. The parametric equation can have one or more Independent variables. Also, a parametric equation for a curve is defined as $x$ and $y$ as a function of a third variable called the Parameters. The cartesian equation for a curve is an equation of a curve in terms of $x$ and $y$ only.