Eliminate the Parameter - Online Tutor, Practice Problems & Exam Prep

Eliminating the parameter in parametric equations involves solving one equation for the parameter and substituting it into the other. For example, given \( x = \sqrt{t} \) and \( y = t - 3 \), solving for \( t \) yields \( t = x^2 \), leading to the rectangular equation \( y = x^2 - 3 \), which represents a parabola. In cases with trigonometric functions, use Pythagorean identities, such as \( \cos^2(t) + \sin^2(t) = 1 \), to eliminate the parameter and derive equations like \( x^2 + \left(\frac{y}{3}\right)^2 = 1 \), representing an ellipse.

Eliminating the Parameter

So in the last couple of videos, we were introduced to graphing parametric equations, and many times that resulted in graphs that looked like familiar equations like 2x minus 3. We're going to explore that connection a little further in this video because some questions will not only have you graph parametric equations, but they'll also ask you to eliminate the parameter. I know that sounds kind of scary, but what I'm going to show you in this video is that all that means is you're going to get rid of the t variable in your parametric equations, and you're going to be left with equations that only involve x and y. We're going to see that it's actually a pretty familiar process we've seen before. Let's go ahead and get started here. Alright? So if I had these two parametric equations and I wanted to eliminate the parameter, all that really means is that when you're given x of t and y of t, you're going to solve one of the equations for t, and then you're going to substitute it into the other. Alright? There are always two ways to do this. You can solve the x equation for t and plug it into y, or vice versa, solve the y equation for t and plug it into x. For example, if I try to solve the y equation for t, because of this t squared over here, if I just wanted to isolate this, I'm basically going to end up with something like y over 3 square rooted is going to end up being t. And if I try to plug that into the x equation, I'm going to be left with an x in terms of the square root of y. That's not always bad, but that's just not a very familiar equation that we're used to. So instead, what I'm going to do is I'm going to go the other way around. I'm going to solve the x equation for t and then plug it into the y. So what I can do here is I can drop the parentheses around t and just end up with x+2 is equal to t and what I can do is whenever I see t inside of the y equation, I'm just going to replace it with x+2. This is really similar to what we did when we did the substitution method for solving systems of equations. And what happens if I plug this expression x+2 into this expression? I just end up with y is equal to 3 times the quantity x plus 2 squared. Notice how I've eliminated the parameter, and all I'm left with is an equation that just involves x's and y's. This is what it means to eliminate the parameter. And, again, 90% of the time, solving x and plugging into y is usually going to be the right move because you'll end up with a more familiar equation this way. Alright? That's all eliminating the parameter actually means.

So, let's go ahead and take a look at our example here because, again, lots of questions will ask you to graph parametric equations and then eliminate the parameter to convert them to a rectangular equation. A rectangular equation is just an equation that involves x's and y's, like this one. Alright? So let's get started. We have x as the square root of t, and then y as t minus 3, and the only restriction for t is that t has to be positive values. And, again, the reason for this is that if t were negative, you'd end up with imaginary numbers because of the square root. Now, again, a lot of times, you're going to have to pick the t's, but in this case, they're already given to you. So, we know how to do this. Let's just graph it first. You have our t values. Remember, these are just going to be inputs that we plug into x and y. Let's get our values. If we do this for t equals 0, x is the square root of t. So at t=0, x=0; the square root of 1 is 1. The square root of 2 is about 1.4, so you can plug that in as a decimal if you'd like. You can use your calculator for this. The square root of 3 is approximately 1.7, and the square root of 4 is 2. What about the y values? For the y values, I'm going to take whatever t is and subtract 3. So this is pretty straightforward. When t is 0, y is negative 3. When t is 1, y is negative 2, so on and so forth. You're going to see a little pattern that emerges here. Okay? So I've got my points, and I can just go ahead and graph these things. So if I graph this, I'm going to get 0, negative 3. That's one of my points. 1, negative 2. That's the second. And then I'm going to skip these other ones over here because they're not sort of clean numbers, but you could totally graph them if you wanted to. The next sort of value that I see is 2, 1, that has a lot of clean numbers. So clearly, we can see this is not going to be the equation of a line. This is actually going to start to curve upwards, kind of like actually like a parabola does. So, it's going to look something like this, and in fact, this graph would continue going onwards like this. Alright? So let's write in the t values. This is where t was equal to 0. This is where t was 1, and this is where t was equal to 4. So that means that the orientation of this plane curve is sort of just going upwards like this, and this graph would keep increasing in this direction. Again, it looks very much like a parabola.

So now let's get to the second part of this question, which is eliminating the parameter to convert it to a rectangular equation. Why does this thing actually end up looking like a parabola? Well, let's get started here. Again, almost always, the simpler way to do this is solving x and then plugging into y. So for example, if I try to solve y for t, I'm going to get something like y plus 3. And when I plug it into this equation, I'm going to get x in terms of the square root of y. Again, not very familiar. So what I'm going to do here is if x of t is equal to t, that means that if I square both sides, that means that x squared is equal to t. And, therefore, what I'm going to do is wherever I see the y equation, y is equal to t minus 3, I'm just going to replace the t with x squared. Right? So all that happens here is this y just becomes x2 minus 3. In other words, this is just x2−3, and there is your equation. So, we've eliminated the parameter, leaving only an equation involving y's and x's. This is a rectangular equation, and it's the equation of a parabola. That's exactly why our shape ended up looking like a parabola. It's because when you eliminate the parameter, you end up with just an equation of a parabola. Now, what's an important note here is why did we only get half of the parabola and not the whole entire thing? Well, that actually has to do with the fact that our t values were restricted for the parametric equations. So a lot of times, what you're going to notice here is that if you're given parametric equations have a restriction for that t interval, the graph is usually going to be a smaller portion of the fuller rectangular equation. y=x2−3 is going to look something like this, but our parametric equation only was a small piece of that, and that's because of the t values. Alright? That's all there is to it, folks. Thanks for watching, and let's get some practice.

Eliminate the parameter to rewrite the following as a rectangular equation.

$x\left(t\right)=2t-1$

$y\left(t\right)=t^5-2$

• A. $y=\left(\frac{x-2}{2}\right)^5-1$
• B. $y=\left(\frac{x-1}{2}\right)^5-2$
• C. $y=\left(\frac{x+1}{2}\right)^5-2$
• D. $y=\left(\frac{x+1}{2}\right)^5-1$

Eliminating the Parameter Example 1

Everyone, in this example, we're going to be taking these two parametric equations that are given to us right here. We're going to be graphing them. We'll be graphing the plain curve, and then we'll be writing their equivalent rectangular equation. Remember, this is just fancy language for eliminating the parameter and getting just an equation involving x's and y's. Let's take care of the graphing part first. We've already seen how to do this. The whole idea is we're going to need a bunch of values for t, x, and y. Let's take a look at our interval here. Our interval is just all real numbers, and if you look at our two equations, they're both actually quadratic in t. Both of them have t² in the equations. So the best numbers to use are always just going to be small ones, like ones that are closer to 0 because the numbers will be nicer. But another thing that will also happen here is because both of these things are squared, then if I plug in the numbers 1, 2, 3, and then I try to plug in the numbers -1, -2, -3, they're going to be the same because they just end up being squared. So, in other words, the best variables or numbers to use for t are going to be something like 0, 1, 2, 3. You don't necessarily have to use these, but I'm just going to use these four numbers here. Remember, the whole idea is that these are inputs. We're going to plug these numbers into both the x and y to get a bunch of coordinates. Let's go ahead and do that.

When I plug in t = 0 into this equation, 0² - 1 is -1. When I plug in t = 1, 1² - 1 is 0. 2² is 4, minus 1 is 3, 3² is 9, minus 1 is 8. Do the same exact thing for the y values. 0² - 2 is -2, 1² - 2 is -1, this becomes 4 minus 2, which is 2, and this becomes 9 minus 2, which is 7. So I've got my 4 points over here, 1 comma -2 and so forth, and I can go ahead and plot them and graph them. 1 comma -2 is actually going to be the point that's, like, right over here. Then I've got 0 comma -1, then I've got 3 comma 2, and then I've finally got 8 comma 7. You might look at these points and think that they should resemble a parabola. So, in fact, when you look at these, you may have actually thought that these were going to look like sort of parabolic shapes, but things don't always turn out the way that they appear to. Alright? So if I graph these things and actually connect them with a straight line, we're going to see that these are not parabolas, but this is in fact the equation of a straight line.

Now if you take a look here, we can figure out the orientation by writing in the t values. In other words, this was when t = 0, this was t = 1, this was t = 2, and this was t = 3. So, if you wanted to indicate the orientation of this line here, this plain curve, this is just going to be in the direction of increasing t. So in other words, it's going to be going this way, like this. Alright? So in other words, these things look like they were going to be parabolas because they're quadratic. When you actually sort of graph these things out, it turns out to be the equation of a straight line.

Alright, that's the plane curve. Now, to understand why we got a straight line, we're going to go ahead and write the equivalent rectangular equation. Remember, this is just x's and y's, and the way we do this is we have to eliminate the parameter. We have to get rid of t, and the way we do this is we solve one equation for t and then plug it into the other one. The easiest thing, if you can, is always going to be to try to solve the x equation for t and then plug it into the y equation because then you'll get y in terms of x instead of x in terms of y. So in other words, when I get x = t² - 1, I just have to solve for t. So this just becomes x + 1 = t² and then I just have to take the square root of both sides. So, in other words, the square root of x + 1 equals t. Now what I do is whenever I see the y equation, y = t² - 2, I'm just going to take that expression that I just found for t and then plug it into the t of the y equation. So this just becomes y = (square root of x + 1)² - 2.

This might look kind of complicated at first, but what happens here is that I have an expression that's square rooted, but then I have it square on the outside. So, basically, those things cancel each other out. And all I get here is that I get x + 1 minus 2, which is if you simplify this, this ends up just being y = x - 1. So, lo and behold, what we found is that we have an equation where there's no t's. It's just x's and y's, and it's not a parabola. It's just the equation of a line with a slope of 1 that goes to the y intercept of -1, which is exactly what we have graphed here. Alright? So that's what happens when you have 2 equations that are parabolic or quadratic in terms of t. You actually end up with just the equation of a line.

That's it for this one, folks. Hopefully, you've had a good time, let me know if you have any questions, and I'll see you in the next video.

Eliminate the Parameter Example 2

Welcome back, everyone. So in this example, we have a couple of parametric equations. We've got x(t)=t−3, y(t)=1t−5. That's a rational equation. And we're going to go ahead and graph the plane curve. And then later on, we're going to write the equivalent rectangular equation. Let's get started here. We know how to do this before. All we need is just a bunch of x and y pairs over here. And, normally, we would just go and start picking a bunch of t values, but they're actually already given to us just to make things a little bit easier. The one thing we're told here is that t can be pretty much anything except for 5. If that seems like it's a really specific number, it's actually just because, if you take a look at this rational equation over here, y=1t−5, you just can't have any t values that are going to make the denominator 0 because then you would end up with something that's undefined. So t can be anything as long as it's not 5. We've got a couple of values here. Let's go ahead and just get started. So we're just going to use these t's as inputs. We're going to plug them into the x and y equations to get our coordinates. Alright?

So for t equals 0, x(t)=t−3, so this just ends up being negative 3. You're just going to subtract 3. So you just go down the x column. So if you go down the x column, this is going to be negative 3. And when t is 3, you just get 0. When t is 4.8, you get 1.8, 2.2, and then you just get 4. Alright? So these are my coordinates. Now for the y values, it's a little bit sort of more challenging or difficult, but it's the same principle. You just plug this into this formula and then just go ahead and evaluate. So if I plug in t equals 0 into 10−5, I just end up with 1−5, which ends up being negative 0.2. All right. So I'm just going to write these as decimals. They'll be easier to visualize on a graph. When t is equal to 0, I'm going to end up getting or sorry, when t is equal to 3, what I end up getting here is 13−5 becomes negative 2. So that's negative one half. So this is just negative 0.5. Keep going here. I'm going to get 14.8−5. So this actually ends up being 1−0.2. I am dividing by a decimal, so I am going to get a higher number, and this ends up being negative 5. So, if I do 5.2, I get the exact same thing except it is going to be 10.2, that's positive. So I just end up getting positive 5. And then finally, 17−5 becomes 0.5. So again, I've written these all as decimals just to help sort of visualize these points.

Let's actually go graph them. So my coordinates are going to be negative 3 comma negative 0.2. So it's going to be something that is really, really, really close to the x-axis all the way on the left. And I've got 0 comma negative 0.5, so that's going to be something over here. I've got 1.8 comma negative 5. So it's going to be close to 2 but a little bit to the left and then all the way down at negative 5. Then I've got positive 2.2 and then positive 5. So, basically, what happens is this point ends up being all the way up here. And then finally, I get 4, comma, 0.5. So I get 4, comma, 0.5 that looks something like this. Notice how there's not really a pattern that emerges. Like, that's clear here. It sort of looks like it's going to be something like this, but that wouldn't resemble any sort of normal function. What we can see here is that you sort of gain this out. If you start getting t values that are really close to 5, you're going to end up getting increasingly larger and larger numbers. So this is actually going to start to look like a rational equation, and that makes sense because the y function is or the y equation is a rational equation. So I'm going to go ahead and sort of draw a sketch, a little curve that sort of resembles what this might look like, and it's going to look something like this. I would just. Yeah. It's let's draw something like this. All right. This is going to look something like that, and it's going to continue on like this. And then I'm going to have values that come in all the way up here, and then I'm going to have that sort of flatten out like this. Or I can clearly see that this t this x equals 2 is going to be a vertical asymptote. Alright? I can't actually cross that line because I can't have t equal 5. Alright? So, this is going to be our parametric equation. And, what happens is if I write out my t values, this is going to be t equals 0. This is equal to t equals this is when t was 3. This is when t was equal to 4.8. This is when t was equal to 5.2. And then finally, this is t equals 7. So, in other words, what we can see here is that the orientation of this plane curve is it sort of goes off to the right these values. Alright? So this would be the orientation of our parametric equation. So why did we get something like this? Well, let's take a look at our equivalent rectangular equation. What we're going to do is we're going to have to eliminate the parameter. Remember, we're going to solve one equation for t and then plug it into the other. Usually, it's easier to solve the x equation and plug it into the y. Not always. But in this case, what's going to happen here is, if you were to try to solve the t equation for the y equation, you're just going to have more work because you're going to have to flip them, and you're going to have to insert that nasty fraction inside of the other. Whereas it's much easier to solve this equation for t and then plug it into the thing that's already rational. Right. So that's what I'm going to do here. So I've got x=t−3, which means that x+3=t. So, therefore, what happens is when I go to my y equation, y=1t−5. Remember, the whole idea here is that you solve for t. So you're going to get some expression for x, and then you're going to plug that in for t to eliminate the parameters. This ends up being 1x+3−5. When you simplify this, this actually ends up being 1x−2. I've eliminated the t's and I get an equation that's just y with x's. And we can see here that this is a rational equation, which is exactly what we expected. And also, what we can see here is that this is a rational equation that's going to have a vertical asymptote at x equals 2. So remember, these types of equations, you're going to get a vertical asymptote at x equals 2. Alright? So it's going to be the opposite of whatever that number is, and that's exactly what we got. We got a rational function in which we couldn't sort of cross that x equals 2 boundary. Alright. That's it for this one, folks. Let me know if you have any questions.

Equations with Trigonometric Functions

So in recent videos, we learned how to eliminate the parameter from parametric equations. We would solve one of the equations for t, plug it into the other, and end up with a familiar equation of a line or a parabola or something like that. Can we use the same strategy when you're given parametric equations involving trig functions? Well, if I were to try this here, I would get something like the inverse cosine of x. And when I plug it into the y equation, I'll get y equals sine of inverse cosine of x. It turns out to be a mess. In fact, that actually just won't work. So it turns out with these types of problems, we'll need a different strategy. I'm going to show you how to do that in this video, and the basic idea here is we're going to rewrite both of the equations and use a familiar Pythagorean identity that we've seen before. Let me walk you through how to do this. Let's get started here. Alright? So, again, the basic idea here is that we were solving one equation for t when we were just eliminating the parameter for these types of equations and plugging into the other.

So if I were given something like, for example, x equals cosine of t and y equals 3 sine of t, how would I go about eliminating the parameter? Well, the idea here is that you're going to solve both of the equations for whatever trig function is inside of them. So for example, here we have cosine, here we have sine, but sometimes you may have sine or tangents and secants and things like that. Alright? So what we're going to do here is we're going to solve both of the equations for whatever trig functions are inside of them. So, for example, this is x equals cosine of t. All I have to do is just get cosine t by itself, but it already is. So in this case, all I have to do is just flip the equation around: cosine t equals x. And I've solved that equation for that whatever trig function. Alright? Same thing for y equals 3 sine of t. What I'm going to do is I'm going to try to get sine t by itself. All I have to do is just move the 3 over to the other side, but I'm going to flip the equation around and say that sine of t is equal to y over 3.

So how does this help us eliminate the parameter? I still have t's inside of both of these equations. Well, now what we're going to do here is we're going to relate our trig functions back to a Pythagorean identity. Remember, these are just the equations that involve the squares of a bunch of trig functions and relating them back to 1. So which one of these 3 Pythagorean identities am I going to use here? Well, in my parametric equation, I have cosine and sine. So it makes sense that I'm going to use the one that involves cosine and sine. So remember, this Pythagorean identity says that the sine squared of some angle plus the cosine squared of some angle is equal to 1. But it actually doesn't have to be an angle. In fact, these variables, the thetas, can actually be anything. It could be x or even t. So what I'm going to do is I'm going to take this first one and I'm going to rewrite it slightly. So what I'm going to do is I'm going to say that cost2 + sint2 = 1. As long as your calculator is in radians mode, no matter what you plug into these equations, you'll always actually just end up getting 1.

So, how does this help us eliminate the parameter? Well, I've got an identity here that says the cosine squared of t plus the sine squared of t is equal to 1. So, what I'm going to have to do here is now that I've solved both of the equations for these trig functions I'm just going to have to square both of them. So, in other words, the cosine squared is just going to be squared and the sine squared is just going to be the expression y over 3 squared. So now, what happens is every time I see cosine squared, I'm just going to replace it with x squared and every time I see the sine squared, I'm just going to replace it with (y/3)² and then this equals 1, and then I just have a plus sign over here. So now we can see this equation over here, x² + (y/3)² = 1, is an equation that no longer involves t, and it's also a pretty familiar equation that we've seen before. This actually ends up being the equation of a familiar conic section, which is an ellipse. You have something like x² + y² with some other numbers here equals 1. So this actually ends up being how you eliminate the parameter from these types of equations that involve trig functions. You have to solve both of the equations for whatever trig functions are in them, and then you have to relate it back to a familiar Pythagorean identity to get rid of that t variable. That's really all there is to it, folks. So let's go ahead and take a look at some practice problems. Thanks for watching.

First eliminate the parameter, then graph the plane curve of the parametric equations.

$x\left(t\right)=2+\cos t$, $y\left(t=-1+\sin t\right)$; $0\le t\le2\pi$

• A. $\left(x-2\right)^2+\left(y+1\right)^2=1$
• B. $\left(x-2\right)^2+\left(y+1\right)^2=1$
• C. $\left(x+2\right)^2+\left(y-1\right)^2=1$
• D. $\left(x+2\right)^2+\left(y-1\right)^2=1$

Equations with Trigonometric Functions Example 3

Everyone, in this example, we've got 2 parametric equations. We've got x(t)=3∙cos(t) and y=2∙sin(t). Now remember, these involve trigonometric functions, so it's always best to rather than first graph it, we're actually just going to first eliminate the parameter so we can get the rectangular equation, and then we can graph the plane curve after we've gotten rid of that parameter t. Alright? Let's go ahead and get started here. Remember, these types of problems when you're eliminating the parameter always boil down to trying to get something in terms of the Pythagorean identity cos2(t) + sin2(t) = 1 . I've got equations here that involve cosine and sine, so I'm just going to have to rewrite those things in terms of the x's and y's. Alright?

So how do we do this? Well, I've got that x=3∙cos(t). I want cos2(t). Let's bring the 3 over to the other side. I'm going to divide by 3, and then I'm basically just going to go ahead and take the square roots. Or sorry. I'm going to square both sides. So in other words, I'm going to get x32=cos2. Alright? So now what happens is every time I see, or wherever I see a cos²(t), I can actually just replace it with x32. Alright? Same thing over here with the y axis or the sine of t. Alright? So that y=2∙sin(t). If I want to rearrange, I've got to divide 2 on both sides, and then I'll just square both sides. So I get y22=sin2t. Alright? So now what happens is wherever I see sin²(t), I can actually just replace it with this expression over here. So if I combine all of that stuff, what I end up with is that x32 + y22 =1. So I've eliminated the parameter. This is going to be my rectangular equation over here. And so what happens is this is my answer. And now that I've eliminated the parameter, now I can go ahead and graph this plain curve because I already know what this thing is going to look like, or I have an idea of what this thing is going to look like because now I have the rectangular equation. If this doesn't look familiar to you, we have x2 + y2, but you have these little numbers here that are inside of the denominators. This is actually the equation of an ellipse. We've seen this before when we talk about conic sections. This is an ellipse because you have these two numbers over here that are not the same that are underneath the x's and y's, and this is going to be an ellipse. It's centered at 0. So it's centered at 0, and basically, the 3 and the 2 are going to make up the semi-major and semi-minor axes or the major-minor axes. So, in other words, because the x is bigger, this is going to be a sort of an oblong oval that's going to be wider in the x than it is in the y. And it's basically to go 3 in the x and then 2 in the y like this. So, basically, what our graph is going to look like is it's going to go from 3 to -3 on the x-axis and then 2 to -2 on the y-axis. And If you connect these things, what you end up getting is you'll end up getting something that looks like this. Alright. So are we actually done here? Well, in most cases, we would be because usually what's going to happen inside of these problems with trigonometric functions is that your parameter t is going to go from 0 to 2π, in which case you're just going to go all the way around the circle. We're not actually quite done here yet because we have to take a look at the restriction of t. t only goes from π all the way to 2π. It doesn't go from 0 all the way to 2π. So what happens here is that this piece, the top half of the ellipse, this would be from 0 to 2π. Oh, sorry. This would be 0 ≤ t ≤ π. And then what happens is the bottom half of the ellipse would be π < t < 2π. Think of just the way that the unit circle works. Right? Once you go all the way to π, that's going to be all the way on the negative x-axis. And then all the way around again would be going back to 2π. So what happens here in this problem is we actually can't graph the top half of the ellipse because that's not inside of the parameter. So we don't have the entire interval from 0 to 2π. We only have from π to 2π. So the only part that you can actually graph here is going to be the bottom half of the ellipse. Alright? That's always an important thing to consider when you are doing these types of problems is that sometimes the t restriction won't actually get you the entire circle or the entire ellipse or whatever shape it is. Alright, folks. So that's it for this one. Let me know if you have any questions.