# 10.6 Parametric equations  (Page 3/6)

 Page 3 / 6

## Eliminating the parameter

In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations.

## Eliminating the parameter from polynomial, exponential, and logarithmic equations

For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ We substitute the resulting expression for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ into the second equation. This gives one equation in $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$

## Eliminating the parameter in polynomials

Given $\text{\hspace{0.17em}}x\left(t\right)={t}^{2}+1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\left(t\right)=2+t,\text{\hspace{0.17em}}$ eliminate the parameter, and write the parametric equations as a Cartesian equation.

We will begin with the equation for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ because the linear equation is easier to solve for $\text{\hspace{0.17em}}t.$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=2+t\hfill \\ y-2=t\hfill \end{array}$

Next, substitute $\text{\hspace{0.17em}}y-2\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ in $\text{\hspace{0.17em}}x\left(t\right).$

The Cartesian form is $\text{\hspace{0.17em}}x={y}^{2}-4y+5.$

Given the equations below, eliminate the parameter and write as a rectangular equation for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function
of $\text{\hspace{0.17em}}x.$

$\begin{array}{l}\\ \begin{array}{l}x\left(t\right)=2{t}^{2}+6\hfill \\ y\left(t\right)=5-t\hfill \end{array}\end{array}$

$y=5-\sqrt{\frac{1}{2}x-3}$

## Eliminating the parameter in exponential equations

Eliminate the parameter and write as a Cartesian equation: $\text{\hspace{0.17em}}x\left(t\right)={e}^{-t}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\left(t\right)=3{e}^{t},\text{\hspace{0.17em}}\text{\hspace{0.17em}}t>0.\text{\hspace{0.17em}}$

Isolate $\text{\hspace{0.17em}}{e}^{t}.\text{\hspace{0.17em}}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x={e}^{-t}\hfill \\ {e}^{t}=\frac{1}{x}\hfill \end{array}$

Substitute the expression into $\text{\hspace{0.17em}}y\left(t\right).$

$\begin{array}{l}y=3{e}^{t}\hfill \\ y=3\left(\frac{1}{x}\right)\hfill \\ y=\frac{3}{x}\hfill \end{array}$

The Cartesian form is $\text{\hspace{0.17em}}y=\frac{3}{x}.$

## Eliminating the parameter in logarithmic equations

Eliminate the parameter and write as a Cartesian equation: $\text{\hspace{0.17em}}x\left(t\right)=\sqrt{t}+2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\left(t\right)=\mathrm{log}\left(t\right).$

Solve the first equation for $\text{\hspace{0.17em}}t.$

Then, substitute the expression for $t$ into the $y$ equation.

$\begin{array}{l}y=\mathrm{log}\left(t\right)\\ y=\mathrm{log}{\left(x-2\right)}^{2}\end{array}$

The Cartesian form is $\text{\hspace{0.17em}}y=\mathrm{log}{\left(x-2\right)}^{2}.$

Eliminate the parameter and write as a rectangular equation .

$\begin{array}{l}\\ \begin{array}{l}x\left(t\right)={t}^{2}\hfill \\ y\left(t\right)=\mathrm{ln}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t>0\hfill \end{array}\end{array}$

$y=\mathrm{ln}\sqrt{x}$

## Eliminating the parameter from trigonometric equations

Eliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem.

First, we use the identities:

$\begin{array}{l}x\left(t\right)=a\mathrm{cos}\text{\hspace{0.17em}}t\\ y\left(t\right)=b\mathrm{sin}\text{\hspace{0.17em}}t\end{array}$

Solving for $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ we have

$\begin{array}{l}\frac{x}{a}=\mathrm{cos}\text{\hspace{0.17em}}t\\ \frac{y}{b}=\mathrm{sin}\text{\hspace{0.17em}}t\end{array}$

Then, use the Pythagorean Theorem:

${\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1$

Substituting gives

${\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t={\left(\frac{x}{a}\right)}^{2}+{\left(\frac{y}{b}\right)}^{2}=1$

## Eliminating the parameter from a pair of trigonometric parametric equations

Eliminate the parameter from the given pair of trigonometric equations where $\text{\hspace{0.17em}}0\le t\le 2\pi \text{\hspace{0.17em}}$ and sketch the graph.

$\begin{array}{l}x\left(t\right)=4\mathrm{cos}\text{\hspace{0.17em}}t\\ y\left(t\right)=3\mathrm{sin}\text{\hspace{0.17em}}t\end{array}$

Solving for $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,$ we have

$\begin{array}{l}\text{\hspace{0.17em}}x=4\mathrm{cos}\text{\hspace{0.17em}}t\hfill \\ \frac{x}{4}=\mathrm{cos}\text{\hspace{0.17em}}t\hfill \\ \text{\hspace{0.17em}}y=3\mathrm{sin}\text{\hspace{0.17em}}t\hfill \\ \frac{y}{3}=\mathrm{sin}\text{\hspace{0.17em}}t\hfill \end{array}$

Next, use the Pythagorean identity and make the substitutions.

$\begin{array}{r}\hfill {\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1\\ \hfill {\left(\frac{x}{4}\right)}^{2}+{\left(\frac{y}{3}\right)}^{2}=1\\ \hfill \frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1\end{array}$

The graph for the equation is shown in [link] .

#### Questions & Answers

bsc F. y algebra and trigonometry pepper 2
Aditi Reply
given that x= 3/5 find sin 3x
Adamu Reply
4
DB
remove any signs and collect terms of -2(8a-3b-c)
Joeval Reply
-16a+6b+2c
Will
is that a real answer
Joeval
(x2-2x+8)-4(x2-3x+5)
Ayush Reply
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
X2-2X+8-4X2+12X-20=0 (X2-4X2)+(-2X+12X)+(-20+8)= 0 -3X2+10X-12=0 3X2-10X+12=0 Use quadratic formula To find the answer answer (5±Root11i)/3
master
Soo sorry (5±Root11* i)/3
master
x2-2x+8-4x2+12x-20 x2-4x2-2x+12x+8-20 -3x2+10x-12 now you can find the answer using quadratic
Mukhtar
explain and give four example of hyperbolic function
Lukman Reply
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
Racelle Reply
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Anurag Reply
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
Jhovie Reply
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
Charmaine Reply
A banana.
Yaona
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
Snalo Reply
what are you up to?
Mark Reply
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
Propessor Reply
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
sita Reply
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
What is algebra
Pearl Reply
algebra is a branch of the mathematics to calculate expressions follow.
Miranda
Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅
Jeffrey
lolll who told you I'm good at it
Miranda
something seems to wispher me to my ear that u are good at it. lol
Jeffrey
lolllll if you say so
Miranda
but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it.
Jeffrey
which grade are you in though
Miranda
oh woww I understand
Miranda
haha. already finished college
Jeffrey
how about you? what grade are you now?
Jeffrey
I'm going to 11grade
Miranda
how come you finished in college and you don't like math though
Miranda
gotta practice, holmie
Steve
if you never use it you won't be able to appreciate it
Steve
I don't know why. But Im trying to like it.
Jeffrey
yes steve. you're right
Jeffrey
so you better
Miranda
what is the solution of the given equation?
Nelson Reply
which equation
Miranda
I dont know. lol
Jeffrey
please where is the equation
Miranda
ask nelson. lol
Jeffrey

### Read also:

#### Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Algebra and trigonometry' conversation and receive update notifications?

 By By By Abishek Devaraj By By OpenStax By Subramanian Divya By Madison Christian By Mackenzie Wilcox By Dindin Secreto By Stephen Voron By Jonathan Long By Brianna Beck