New Zealand
Level 8 - NCEA Level 3

Rectangular Equations for Parametric equation

Lesson

The rectangular equation corresponding to a line or a curve, which is given parametrically as $x=g\left(t\right),y=h\left(t\right)$x=g(t),y=h(t) can be found by eliminating $t$t from the equations.

For example, the function $x=2t,y=t^2$x=2t,y=t2 for real $t$t, shows $t=\frac{x}{2}$t=x2 and thus $y=t^2=\left(\frac{x}{2}\right)^2=\frac{1}{4}x^2$y=t2=(x2)2=14x2. This graph shows how $t$t changes for various values of $x$x:

The elimination of $t$t from the parametric equations can require a knowledge of certain established identities, such as the Pythagorean identities $\sin^2\theta+\cos^2\theta=1$sin2θ+cos2θ=1$\sec^2\theta=1+\tan^2\theta$sec2θ=1+tan2θ  and $\csc^2\theta=1+\cot^2\theta$csc2θ=1+cot2θ.   There are also shortcuts that can help you avoid unnecessary algebraic complexity.

Here are four more examples, along with their graphs.

Example 1

$x=t-4$x=t4, $y=\frac{12}{t^2-8t+16}$y=12t28t+16

Without thinking carefully before we proceed, we might set $t=x+4$t=x+4 and then substitute in to the second equation so that $y=\frac{12}{\left(x+4\right)^2-8\left(x+4\right)+16}$y=12(x+4)28(x+4)+16. Simplification of this becomes cumbersome.

However, noticing first that $t^2-8t+16=\left(t-4\right)^2$t28t+16=(t4)2, we simply write:

 $y$y $=$= $\frac{12}{t^2-8t+16}$12t2−8t+16​ $=$= $\frac{12}{\left(t-4\right)^2}$12(t−4)2​ $=$= $\frac{12}{x^2}$12x2​

The graph shows a couple of values of $t$t adjacent to the corresponding coordinates:

Example 2

$x=12\sin t,y=-12\cos t$x=12sint,y=12cost for $t$t in $\left[0,2\pi\right]$[0,2π].

Here, instead of isolating $t$t from either equation, we square both sides of each equation and add them together, so that:

 $x^2$x2 $=$= $144\sin^2t$144sin2t     $(1)$(1) $y^2$y2 $=$= $144\cos^2t$144cos2t     $(2)$(2) $\therefore$∴   $x^2+y^2$x2+y2 $=$= $144\sin^2t+144\cos^2t$144sin2t+144cos2t $=$= $144\left(\sin^2t+\cos^2t\right)$144(sin2t+cos2t) $\therefore$∴   $x^2+y^2$x2+y2 $=$= $144$144

This is a circle, centre origin and radius $12$12. The same result would have occurred had the parametric equations been given as $x=12\sin t,y=12\cos t$x=12sint,y=12cost, with the negative sign in the second equation omitted.

The interesting difference between the two is the starting position $t=0$t=0. In the original question, the circle is drawn from the point $\left(0,-12\right)$(0,12), whereas in the case with the negative sign missing, the circle is drawn from $\left(0,12\right)$(0,12).

Example 3

Sketch the curve given by $x=\ln t,y=\frac{1}{t}$x=lnt,y=1t and within the restricted domain $x>0$x>0.

Note first that for $t$t to be defined, it must be positive, but this restriction may not effect the domain of the function itself. However, in this instance, an additional restriction of $x>0$x>0 has been placed on the function, and so the the graph has the domain given by $\left(0,\infty\right]$(0,].

From $x=\ln t$x=lnt we have $t=e^x$t=ex, and therefore $y=\frac{1}{e^x}=e^{-x}$y=1ex=ex.

So for example, at $x=\ln e=1$x=lne=1, since $t=e$t=e we have $y=\frac{1}{e}$y=1e. Similarly, at $x=\ln e^2=2$x=lne2=2, $y=\frac{1}{e^2}$y=1e2.

Example 4

$x=5+3\cos t,y=3+5\sin t$x=5+3cost,y=3+5sint for $t$t in $\left[0,2\pi\right]$[0,2π].

From the first equation, we have $\frac{x-5}{3}=\cos t$x53=cost and from the second equation we have $\frac{y-3}{5}=\sin t$y35=sint. By adding the squares of both sides of these equations, we have:

 $\frac{\left(x-5\right)^2}{9}+\frac{\left(y-3\right)^2}{25}$(x−5)29​+(y−3)225​ $=$= $\cos^2t+\sin^2t$cos2t+sin2t $\therefore$∴     $\frac{\left(x-5\right)^2}{9}+\frac{\left(y-3\right)^2}{25}$(x−5)29​+(y−3)225​ $=$= $1$1

This is the equation of an ellipse whose centre is located at the point $\left(5,3\right)$(5,3) and whose major and minor axis lengths are $10$10 and $6$6 respectively.

Worked Questions

Question 1

Consider the curve defined by $x=t+3$x=t+3, $y=\frac{2}{t+3}$y=2t+3, for $t\ne-3$t3.

1. Find the rectangular equation for the curve.

2. Which of the following is the graph of the curve?

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A

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B

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C

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D

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A

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B

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C

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D

Question 2

Consider the curve defined by $x=-2+5\cos t$x=2+5cost, $y=-3+5\sin t$y=3+5sint, for $t$t in [$0$0, $2\pi$2π].

1. Find the rectangular equation for the curve.

2. Graph the curve.

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Question 3

Consider the curve defined by $x=4+3\cos t$x=4+3cost, $y=-1+5\sin t$y=1+5sint, for $t$t in [$0$0, $2\pi$2π].

1. Find the rectangular equation for the curve.

2. Which of the following is the graph of the curve?

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A

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B

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C

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D

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A

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B

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C

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D

Outcomes

M8-1

Apply the geometry of conic sections

91573

Apply the geometry of conic sections in solving problems