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.
$x=t-4$x=t−4, $y=\frac{12}{t^2-8t+16}$y=12t2−8t+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)2−8(x+4)+16. Simplification of this becomes cumbersome.
However, noticing first that $t^2-8t+16=\left(t-4\right)^2$t2−8t+16=(t−4)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:
$x=12\sin t,y=-12\cos t$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).
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=e−x.
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.
$x=5+3\cos t,y=3+5\sin t$x=5+3cost,y=3+5sint for $t$t in $\left[0,2\pi\right]$[0,2π].
From the first equation, we have $\frac{x-5}{3}=\cos t$x−53=cost and from the second equation we have $\frac{y-3}{5}=\sin t$y−35=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.
Consider the curve defined by $x=t+3$x=t+3, $y=\frac{2}{t+3}$y=2t+3, for $t\ne-3$t≠−3.
Find the rectangular equation for the curve.
Which of the following is the graph of the curve?
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].
Find the rectangular equation for the curve.
Graph the curve.
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].
Find the rectangular equation for the curve.
Which of the following is the graph of the curve?
Apply the geometry of conic sections
Apply the geometry of conic sections in solving problems