Eliminating and Choosing Parameters

What this move does

To turn parametric equations ${x = f (t) y = g (t)$ into a Cartesian equation $F (x, y) = 0$ , the key is to eliminate $t$ :

eliminate t ⟹ F (x, y) = 0 + the ranges of x, y

Choose a way to eliminate: if you can solve for $t$ , substitute; if there are $sin t$ and $cos t$ , square and add or use a double-angle formula; if the two equations have matching structure, add, subtract or multiply them to get something free of $t$ .
Eliminate $t$ and tidy up into $F (x, y) = 0$ .
Look back: use the range of the parameter to find the range of $x$ (or $y$ ), and write it next to the equation. Skip this step and the Cartesian equation often has “an extra piece” the original curve does not have.

Going the other way, there are also two routes to graphing parametric equations: make a table of $t$ values and plot points; or eliminate first, recognize the curve, and then cut it down to the right range.

Predict first

What figure is described by $x = cos^{2} t, y = sin^{2} t$ ?

The line

x + y = 1

A line segmentA rayA circle

Choose an answer first; then you can see the result.

Work through one

Example: turn x = sin t, y = cos 2t into a Cartesian equation

Step 1 / 4

Trace it first
$t$ goes from $0$ to $2 π$ . The point moves back and forth along an arc: $x$ grows from $0$ to $1$ , falls back to $- 1$ , then returns to $0$ .

${x = sin t y = cos 2 t$
Choose the tool: the double-angle formula
The graph of the Cartesian equation
Look back at the range

Choosing the parameter

The same curve has many sets of parametric equations. With the angle $θ$ as parameter, the circle $x^{2} + y^{2} = 4$ is $x = 2 cos θ, y = 2 sin θ$ ; put $t = tan \frac{θ}{2}$ and it becomes a rational parametrization

x = \frac{2 ( 1 - t ^{2} )}{1 + t ^{2}}, y = \frac{4 t}{1 + t ^{2}} .

A rational parameter is convenient for algebra (for example, for finding the rational points on the circle), but the price is one lost point:

Pull

t

from

- 12

12

. The larger

t

gets, the closer the point comes to

(- 2, 0)

, but it never gets there.

The rule for choosing a parameter: it must determine the moving point, and it should have a geometric meaning that makes the conditions easy to write down. Common choices: an angle (for a circle, or the eccentric angle of an ellipse), a slope (for lines through a fixed point), the signed distance $t$ along a line (the star of the next stop), and time (for motion problems).

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: The curve is the same before and after elimination

$x = t + \frac{1}{t}, y = t - \frac{1}{t}$ : squaring and subtracting gives $x^{2} - y^{2} = 4$ . For $t > 0$ the point covers the whole right branch, for $t < 0$ the whole left branch: not a single point of the hyperbola is missing.

Practice

Exercise 1Basic

Write $x = t - 1, y = t^{2} + 2 t$ in the form $y = F (x)$ . What is $F (x)$ ?

Exercise 2Basic

Eliminating the parameter from $x = sin θ + cos θ, y = sin θ cos θ$ gives $y = \frac{x ^{2} - 1}{2}$ . If the range of $x$ is $[- m, m]$ , find $m^{2}$ .

Exercise 3Advanced

The curve $x = \frac{1 - t ^{2}}{1 + t ^{2}}, y = \frac{2 t}{1 + t ^{2}} (t \in R)$ is the unit circle with one point removed. What is the $x$ -coordinate of the missing point?

Exercise 4Challenge

On the curve $x = t + \frac{1}{t}, y = t^{2} + \frac{1}{t ^{2}} (t \neq = 0)$ , what is the minimum value of $y$ ?

What this move does

Work through one

Example: turn x = sin t, y = cos 2t into a Cartesian equation

Choosing the parameter

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Handing in as soon as the parameter is gone, forgetting the ranges of x and y

Changing the parameter and quietly losing a point