NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Alternate Forms of Parametric Equations
Lesson

Given the parametric form of a relation $x=g\left(t\right),y=h\left(t\right)$x=g(t),y=h(t) we can find the relation's unique rectangular form by making $t$t the subject in $x=g\left(t\right)$x=g(t) and substituting this into $y=h\left(t\right)$y=h(t).

So for example, for the function $x=2t,y=2t^2$x=2t,y=2t2, we find from the first equation that $t=\frac{x}{2}$t=x2 and thus the function is uniquely described by the equation $y=2t^2=2\left(\frac{x}{2}\right)^2=\frac{1}{2}x^2$y=2t2=2(x2)2=12x2.

However, going the other way - from rectangular to parametric - will result in a form that depends on the choice of $g\left(t\right)$g(t).

To explain this, suppose for the function $y=\frac{1}{2}x^2$y=12x2, we set $x=3t$x=3t. Then upon substituting into the rectangular equation we have:

 $y$y $=$= $\frac{1}{2}x^2$12​x2 $=$= $\frac{1}{2}\left(3t\right)^2$12​(3t)2 $=$= $y=\frac{9t^2}{2}$y=9t22​

This is a new parametric form for the same function $y=\frac{1}{2}x^2$y=12x2. Thus both parametric forms $x=2t,y=2t^2$x=2t,y=2t2 and $x=3t,y=\frac{9t^2}{2}$x=3t,y=9t22 result in the same rectangular equation.

The explanation for this lies in the way we arbitrarily define $x$x as a function of  $t$t.

Think about the point $\left(12,72\right)$(12,72) on  $y=\frac{1}{2}x^2$y=12x2. Using $x=2t,y=2t^2$x=2t,y=2t2 we have $2t=12$2t=12 and so $t=6$t=6, and thus $y=2\left(6\right)^2=72$y=2(6)2=72.

On the other hand, using $x=3t,y=\frac{9t^2}{2}$x=3t,y=9t22 we have $3t=12$3t=12 and so $t=4$t=4 and thus $y=\frac{9\left(4\right)^2}{2}=72$y=9(4)22=72.

Both parametric forms are thus legitimate expressions for the one function. There are any number of parametric expressions for each rectangular equation.

##### Example 1

Derive two parametric expressions for the line given by $y=2x-8$y=2x8.

We have plenty of choices. Suppose we set $x=t+4$x=t+4. Then $y=2x-8=2\left(t+4\right)-8=2t$y=2x8=2(t+4)8=2t. This means that the parametric equations $x=t+4,y=2t$x=t+4,y=2t is a legitimate form.

Now suppose $x=2t$x=2t. Then $y=2\left(2t\right)-8=4t-8$y=2(2t)8=4t8. Again, $x=2t,y=4t-8$x=2t,y=4t8 is another legitimate form.

##### Example 2

Use the parameter $t=\frac{1}{x}-5$t=1x5 to find a set of parametric equations that represent the line $x=\frac{y-1}{5}$x=y15.

If $t=\frac{1}{x}-5$t=1x5, we can rearrange to show that $x=\frac{1}{t+5}$x=1t+5. Substituting this into the rectangular form, we have:

 $x$x $=$= $\frac{y-1}{5}$y−15​ $\frac{1}{t+5}$1t+5​ $=$= $\frac{y-1}{5}$y−15​ $\frac{5}{t+5}$5t+5​ $=$= $y-1$y−1 $\therefore$∴    $y$y $=$= $1+\frac{5}{t+5}$1+5t+5​ $=$= $\frac{t+5}{t+5}+\frac{5}{t+5}$t+5t+5​+5t+5​ $=$= $\frac{t+10}{t+5}$t+10t+5​

Hence, a parametric representation for the function is given by $x=\frac{1}{t+5},y=\frac{t+10}{t+5}$x=1t+5,y=t+10t+5.

##### Example 3

The rectangular (or cartesian) equation for a circle, centre origin and radius $5$5 is given by $x^2+y^2=25$x2+y2=25. Show that $x=5\cos t,y=5\sin t$x=5cost,y=5sint are parametric equations that represent it.

Then, by substitution:

 $x^2+y^2$x2+y2 $=$= $\left(5\cos t\right)^2+\left(5\sin t\right)^2$(5cost)2+(5sint)2 $=$= $25\left(\sin^2t+\cos^2t\right)$25(sin2t+cos2t) $=$= $25\times1$25×1 $=$= $25$25

Hence, $x=5\cos t,y=5\sin t$x=5cost,y=5sint is a parametric representation of the circle.

Note that this is not unique. For example $x=5\cos2t,y=5\sin2t$x=5cos2t,y=5sin2t is another legitimate representation.

##### Example 4

Use $x=6\sec t$x=6sect to determine a parametric representation of the hyperbola given by $\frac{x^2}{36}-\frac{y^2}{49}=1$x236y249=1.

Substituting, we have:

 $\frac{\left(6\sec t\right)^2}{36}-\frac{y^2}{49}$(6sect)236​−y249​ $=$= $1$1 $\sec^2t-1$sec2t−1 $=$= $\frac{y^2}{49}$y249​ $\tan^2t$tan2t $=$= $\frac{y^2}{49}$y249​ $y^2$y2 $=$= $49\tan^2t$49tan2t $\therefore$∴    $y$y $=$= $7\tan t$7tant

Therefore $x=6\sec t,y=7\tan t$x=6sect,y=7tant are parametric representations of the hyperbola.

#### Worked Questions

##### Question 1

We want to find parametric representations for the line $y=5x-2$y=5x2.

1. If we let one parametric equation be $x=t$x=t, what would the parametric equation for $y$y be?

2. If we let one parametric equation be $x=t+2$x=t+2, what would the parametric equation for $y$y be?

##### Question 2

Use the parameter $t=\frac{1}{x}-3$t=1x3 to find a set of parametric equations that represent the curve $x=\frac{1}{y+3}$x=1y+3.

##### Question 3

Consider the hyperbola with equation $\frac{x^2}{4}-\frac{y^2}{9}=1$x24y29=1.

If $x$x is represented parametrically by $x=2\sec t$x=2sect, state the parametric representation of $y$y.

### Outcomes

#### M8-1

Apply the geometry of conic sections

#### 91573

Apply the geometry of conic sections in solving problems