Eliminating the parameter
A pair of parametric equations traces a path one moment at a time. Which curve is that path, written as a single equation in x and y?
Where you are 10 in Polar coordinates and vectors
When you reach for it
A pair x = f(t), y = g(t) (see Parametric equations) gives the curve one point at a time. To say which curve it is, a line, a parabola, a circle, an ellipse or a hyperbola, you want an equation in x and y alone, the kind whose graph you already know. Getting rid of t to reach that equation is eliminating the parameter.
The trade has two sides. The single equation names the shape, but it forgets the starting point, the direction and the timing. And it can describe more than the path: the equation y = 4 - x^2 is a whole parabola, while a parametric pair may trace only half of it. So the finished answer is always an equation together with the part of its graph that is traced and the direction of travel.
The procedure
Solve one equation for t and substitute. At every point of the curve, both equations hold with the same value of t. If one of them can be solved for t with a single answer, as in Literal equations: solving for one variable, that expression is the t of the point, and putting it into the other equation leaves an equation in x and y that every point of the curve satisfies. Choose the equation that gives one answer. From x = 3t - 2 there is only \displaystyle t = \frac{x + 2}{3}, while y = t^2 + 1 gives \displaystyle t = \pm\sqrt{y - 1}, two candidates, and picking one silently drops half the curve.
Use an identity. When the coordinates are trigonometric, solving for t would bring in inverse functions with restricted outputs. Instead, isolate the trigonometric functions and feed them into an identity that holds for every t, such as the Pythagorean identities
The first turns cosine and sine coordinates into a circle or an ellipse; the second turns secant and tangent coordinates into a hyperbola. Other identities work the same way, for instance the double angle formula \cos 2t = 2\cos^2 t - 1.
State the part traced and the direction. Both moves prove one direction only: every point of the curve lies on the graph of the new equation. They do not prove that every point of that graph is reached. The interval for t and the ranges of f and g decide that. A square root is never negative, t^2 - 1 is never below -1, and \cos t never leaves [-1, 1]. So find the set of values that x (or y) actually takes, write it beside the equation, and read the orientation from a coordinate that always increases or always decreases with t, or from a few points in order of t. If a coordinate takes the same value at two different values of t, some points may be visited twice.
Worked examples
A line
Eliminate the parameter from x = 3t - 2, y = 5 - 2t, first for all real t, then for 0 \le t \le 2.
SolutionThe x-equation has one solution for t: add 2 and divide by 3 to get \displaystyle t = \frac{x + 2}{3}. Substitute into the y-equation:
Multiplying by 3 gives the tidier 2x + 3y = 11. Check with the original pair: 2(3t - 2) + 3(5 - 2t) = 6t - 4 + 15 - 6t = 11 for every t.
For all real t, x = 3t - 2 takes every real value, so the whole line is traced. Since x increases with t, it runs left to right (and downward, since the slope is negative). For 0 \le t \le 2, x runs from -2 to 4: the curve is the segment of 2x + 3y = 11 with -2 \le x \le 4, from (-2, 5) to (4, 1).
A parabola traced backward
Eliminate the parameter from x = 2 - t, y = t^2 + 1, for all real t, and describe the orientation.
SolutionSolve the linear equation, not the quadratic one: t = 2 - x, a single answer. Then
Since (2 - x)^2 = (x - 2)^2, this is y = (x - 2)^2 + 1, an upward parabola with vertex (2, 1), reached at t = 0. As t runs through all real numbers, x = 2 - t takes every real value, so the whole parabola is traced. But x decreases as t increases: the point travels from right to left, down to the vertex and up the other side. The equation y = x^2 - 4x + 5 shows none of this; the pair x = t, y = t^2 - 4t + 5 draws the same picture left to right.
Half a parabola
Eliminate the parameter from \displaystyle x = \sqrt{t}, y = 4 - t, t \ge 0.
SolutionSquare the first equation: t = x^2. Squaring is safe here in one direction only. Every point of the curve has \displaystyle x = \sqrt{t}, so x^2 = t; but x^2 = t also allows negative x, which \displaystyle \sqrt{t} never gives. Substituting,
The restriction is the answer’s other half. As t runs from 0 upward, \displaystyle x = \sqrt{t} starts at 0 and grows without bound, so the curve is the right half of the downward parabola, starting at (0, 4) and moving right and down through (1, 3) at t = 1 and (2, 0) at t = 4.
A line that is really a ray
Nana eliminates the parameter from x = t^2 - 1, y = 2t^2 + 3, t any real number. She writes t^2 = x + 1, so y = 2(x + 1) + 3 = 2x + 5, and calls the curve a line. Describe the curve correctly.
SolutionHer algebra is right: every point satisfies y = 2x + 5. But t^2 \ge 0, so x = t^2 - 1 \ge -1. The curve is the ray
starting at (-1, 3). The motion is not one pass. As t rises from very negative values to 0, t^2 shrinks, so the point comes in along the ray to (-1, 3); as t goes on from 0, it travels back out along the same ray. Since t and -t give the same point, every point is visited twice except the endpoint (-1, 3), which is visited once.
Half an ellipse with the Pythagorean identity
Eliminate the parameter from x = 1 + 4\cos t, y = -2 + 3\sin t, 0 \le t \le \pi.
SolutionIsolate the trigonometric functions: \displaystyle \cos t = \frac{x - 1}{4} and \displaystyle \sin t = \frac{y + 2}{3}. Since \cos^2 t + \sin^2 t = 1 for every t,
an ellipse centered at (1, -2) that reaches 4 units left and right of the center and 3 units up and down.
Now the part traced. On 0 \le t \le \pi, \sin t \ge 0, so y \ge -2: only the upper half. At t = 0 the point is (5, -2), at \displaystyle t = \frac{\pi}{2} the top (1, 1), and at t = \pi the point (-3, -2). The upper half is traced once, counterclockwise, from right to left.
A branch of a hyperbola
Eliminate the parameter from x = 2\sec t, y = 3\tan t, \displaystyle -\frac{\pi}{2} < t < \frac{\pi}{2}.
SolutionHere \displaystyle \sec t = \frac{x}{2} and \displaystyle \tan t = \frac{y}{3}, and the identity \sec^2 t - \tan^2 t = 1 gives
a hyperbola with vertices (\pm 2, 0) and asymptotes \displaystyle y = \pm\frac{3}{2}x. The identity holds wherever \sec t and \tan t are defined, which is why the interval stops short of \displaystyle \pm\frac{\pi}{2}.
On this interval \cos t lies in (0, 1], so \displaystyle \sec t = \frac{1}{\cos t} \ge 1 and x \ge 2. Only the right branch is traced. Meanwhile \tan t increases through every real value, so y does too: the point comes up from the lower right, passes the vertex (2, 0) at t = 0, and climbs to the upper right, closer and closer to the asymptotes.
Going the other way
Florentino wants parametric equations for the piece of y = 6 - x^2 from (-1, 5) to (2, 2), traced left to right. Then he wants the same piece traced right to left.
SolutionAny graph y = F(x) has the parametrization x = t, y = F(t): the parameter simply is the x-coordinate. The piece runs over -1 \le x \le 2, so
with x increasing, left to right. To reverse, let x = -t, so y = 6 - (-t)^2 = 6 - t^2. Now x must run from 2 down to -1, so t runs from -2 to 1:
At t = -2 the point is (2, 2) and at t = 1 it is (-1, 5). Eliminating the parameter from either pair returns y = 6 - x^2 with -1 \le x \le 2, which is the check.
Boundaries
- Elimination gives an equation that every point of the curve satisfies. The converse needs a separate argument: the parametric curve may be only a piece of the graph (Examples 3, 4, 5 and 6).
- Solving for t must give one answer. Taking \displaystyle t = \sqrt{x} from x = t^2 assumes t \ge 0 and silently keeps only part of the curve (Exercise 21).
- An identity applies only where its functions are defined: \sec^2 t - \tan^2 t = 1 fails at \displaystyle t = \frac{\pi}{2} + k\pi, k any integer, where \sec t and \tan t are undefined and the left side has no value.
- The equation in x and y loses the orientation, the initial point, the speed, and how many times each point is visited. Record them separately when they matter.
- Not every pair can be eliminated with the functions of this course. For x = t + \sin t there is no way to write t in terms of x with them, yet the pair still traces a definite curve.
Where this goes next
The conics met here, ellipses and hyperbolas, get their own lessons, where the equations from Examples 5 and 6 are read for vertices, foci and asymptotes. Converting an equation between the two systems does the same kind of rewriting for polar equations. In calculus, Derivatives of parametric curves and Arc length and speed on parametric and polar curves work with the parameter directly, because slope, speed and length depend on the motion, not just the shape.
Exercises
Solving for the parameter
- Eliminate the parameter from x = 4t + 1, y = 3 - t, t any real number. Give the orientation.
- Eliminate the parameter from x = 2 - 3t, y = 6t + 1, 0 \le t \le 1, and describe the curve.
- Eliminate the parameter from x = t - 3, y = t^2 + 2t, t any real number.
- Eliminate the parameter from x = t + 2, \displaystyle y = \frac{1}{t}, t \ne 0. Which x is never reached?
- Eliminate the parameter from x = e^t, y = e^{2t} + 1, t any real number (see The number e and the natural exponential function), and state the part traced.
- Eliminate the parameter from \displaystyle x = \sqrt{t - 1}, y = 2t, t \ge 1. Give the initial point and the orientation.
- Eliminate the parameter from x = t^2, y = t - 1, t any real number, writing x in terms of y. Describe the curve and its orientation.
- Eliminate the parameter from x = 3t^2, y = t^2 - 2, t any real number, and describe the motion.
Using identities
- Eliminate the parameter from x = -1 + 2\cos t, y = 4 + 2\sin t, 0 \le t \le 2\pi.
- Eliminate the parameter from x = 3\cos t, y = 2\sin t, \displaystyle \frac{\pi}{2} \le t \le \frac{3\pi}{2}, and state which part is traced and in which direction.
- Eliminate the parameter from x = 2 + \sin t, y = 4\cos t, 0 \le t \le 2\pi, and give the orientation.
- Eliminate the parameter from x = 4\sec t, y = 2\tan t, \displaystyle -\frac{\pi}{2} < t < \frac{\pi}{2}, and describe the part traced.
- Eliminate the parameter from x = \tan t, y = \sec t, \displaystyle -\frac{\pi}{2} < t < \frac{\pi}{2}, and describe the part traced.
- Use \cos 2t = 2\cos^2 t - 1 to eliminate the parameter from x = \cos t, y = \cos 2t, 0 \le t \le \pi. Give the part traced and the orientation.
Going the other way, and uses
- Wenceslao wants the piece of y = x^2 - 5x from (0, 0) to (5, 0) as a parametric curve. Write one traced left to right and one traced right to left, each with 0 \le t \le 5.
- A ball follows x = 30t, y = 40t - 16t^2, 0 \le t \le 2.5 (feet, seconds). Eliminate the parameter, give the interval of x, and find the highest point from the equation in x.
- Nana eliminates the parameter from x = 2\cos^2 t, y = 3\sin^2 t, 0 \le t \le \pi, and gets the line \displaystyle \frac{x}{2} + \frac{y}{3} = 1. Check her equation and describe the curve exactly.
Reasoning
- Explain why eliminating the parameter proves that every point of the curve lies on the graph of the new equation, but not that every point of that graph lies on the curve.
- Show that x = 2t, y = 4t^2 and x = -2t, y = 4t^2, for all real t, give the same equation. What does the equation fail to record?
- Which part of \displaystyle \frac{x^2}{4} - \frac{y^2}{9} = 1 do x = 2\sec t, y = 3\tan t trace for \displaystyle \frac{\pi}{2} < t < \frac{3\pi}{2}, and in which direction?
- For x = t^2, y = t + 1, t any real number, Florentino writes \displaystyle t = \sqrt{x} and gets \displaystyle y = \sqrt{x} + 1. What part of the curve does his equation miss, and what is a correct equation?
- The pair x = t, \displaystyle y = \sqrt{9 - t^2}, -3 \le t \le 3, satisfies x^2 + y^2 = 9. Does it trace the whole circle? Give a pair that does.
Answers
- \displaystyle t = \frac{x - 1}{4}, so \displaystyle y = 3 - \frac{x - 1}{4} = -\frac{1}{4}x + \frac{13}{4}. The whole line, traced left to right since x increases with t.
- \displaystyle t = \frac{2 - x}{3}, so y = 2(2 - x) + 1 = 5 - 2x. As t runs from 0 to 1, x runs from 2 to -1: the segment of y = 5 - 2x with -1 \le x \le 2, from (2, 1) to (-1, 7).
- t = x + 3, so y = (x + 3)^2 + 2(x + 3) = x^2 + 8x + 15. The whole parabola, left to right.
- t = x - 2, so \displaystyle y = \frac{1}{x - 2}. Since t \ne 0, x \ne 2: both branches, each traced left to right (t < 0 gives the lower left branch, t > 0 the upper right).
- e^{2t} = (e^t)^2 = x^2, so y = x^2 + 1. Since e^t > 0 takes every positive value, the part traced is x > 0, the right half without the vertex (0, 1), left to right.
- x^2 = t - 1 with x \ge 0, so t = x^2 + 1 and y = 2x^2 + 2, x \ge 0. Initial point (0, 2) at t = 1; the point moves right and up.
- t = y + 1, so x = (y + 1)^2: a parabola opening to the right with vertex (0, -1). Since y = t - 1 takes every value and increases, the whole parabola is traced upward.
- \displaystyle t^2 = \frac{x}{3}, so \displaystyle y = \frac{x}{3} - 2 with x \ge 0: the ray from (0, -2). As t rises to 0 the point moves in to (0, -2), then back out along the same ray.
- \displaystyle \cos t = \frac{x + 1}{2}, \displaystyle \sin t = \frac{y - 4}{2}, so (x + 1)^2 + (y - 4)^2 = 4: the whole circle with center (-1, 4) and radius 2, once counterclockwise from (1, 4).
- \displaystyle \frac{x^2}{9} + \frac{y^2}{4} = 1. On this interval \cos t \le 0, so x \le 0: the left half, from (0, 2) through (-3, 0) at t = \pi to (0, -2), counterclockwise.
- \sin t = x - 2, \displaystyle \cos t = \frac{y}{4}, so \displaystyle (x - 2)^2 + \frac{y^2}{16} = 1, the whole ellipse with center (2, 0). At t = 0 it is at the top (2, 4) and at \displaystyle t = \frac{\pi}{2} at (3, 0), so it runs clockwise.
- \displaystyle \frac{x^2}{16} - \frac{y^2}{4} = 1. Here \sec t \ge 1, so x \ge 4: the right branch, traced upward since y = 2\tan t increases.
- \sec^2 t - \tan^2 t = 1 gives y^2 - x^2 = 1. Here \sec t \ge 1, so y \ge 1: the upper branch, traced left to right since x = \tan t increases.
- y = 2x^2 - 1. As t runs from 0 to \pi, x = \cos t falls from 1 to -1, so the part with -1 \le x \le 1 is traced right to left, from (1, 1) through the vertex (0, -1) at \displaystyle t = \frac{\pi}{2} to (-1, 1).
- Left to right: x = t, y = t^2 - 5t. Right to left: x = 5 - t, y = (5 - t)^2 - 5(5 - t) = t^2 - 5t. The y-formula is the same because the parabola is symmetric about x = 2.5.
- \displaystyle t = \frac{x}{30}, so \displaystyle y = \frac{4}{3}x - \frac{4}{225}x^2, 0 \le x \le 75. The vertex is at \displaystyle x = \frac{4/3}{2 \cdot 4/225} = 37.5, where y = 50 - 25 = 25: the highest point is (37.5, 25).
- \displaystyle \cos^2 t = \frac{x}{2} and \displaystyle \sin^2 t = \frac{y}{3} add to 1, so the equation holds. But 0 \le \cos^2 t \le 1, so 0 \le x \le 2: only the segment from (2, 0) to (0, 3). The point goes from (2, 0) at t = 0 to (0, 3) at \displaystyle t = \frac{\pi}{2} and back to (2, 0) at t = \pi.
- Each step (substituting the t of a point, or applying an identity true for every t) uses facts that hold at every point of the curve, so each point satisfies the result. Nothing in those steps says that a given solution (x, y) comes from some allowed t; the range of the parametric functions decides that, as in Example 3, where points with x < 0 solve y = 4 - x^2 but are never reached.
- In both, \displaystyle t = \pm\frac{x}{2} and y = 4t^2 = x^2. Both trace the whole parabola y = x^2, the first left to right and the second right to left. The equation records neither direction, nor where the point is at a given t.
- There \cos t < 0, so \sec t \le -1 and x \le -2: the left branch. Since \tan t increases through every real value, the branch is traced upward, through the vertex (-2, 0) at t = \pi.
- \displaystyle t = \sqrt{x} holds only for t \ge 0, so his equation gives only y \ge 1, the upper half. Correct: t = y - 1, so x = (y - 1)^2, the whole parabola opening to the right.
- No. \displaystyle y = \sqrt{9 - t^2} \ge 0, so only the upper half is traced, left to right. The pair x = 3\cos t, y = 3\sin t, 0 \le t \le 2\pi traces the whole circle.