P4.3 - Coordinate geometry in the (x, y) plane

Syllabus
2019
Topic
P4.3
Level
A2

Move between a parameter and a Cartesian curve

Parametric equations give both coordinates in terms of a third variable: x=f(t)x=f(t) and y=g(t)y=g(t). Each permitted value of tt produces a point (x,y)(x,y); changing tt traces the curve. A Cartesian equation removes tt and links xx directly to yy.

To convert to Cartesian form, make tt or a simple function of tt the subject of one equation and substitute into the other. For trigonometric parameters, isolate sint\sin t and cost\cos t and use sin2t+cos2t=1\sin^2t+\cos^2t=1. Finally translate the stated range of tt into restrictions on xx or yy.

For x=(t+3)/(t+1)x=(t+3)/(t+1) and y=2/(t+1)y=2/(t+1) with t>0t>0, notice that x=1+2/(t+1)=1+yx=1+2/(t+1)=1+y. Hence y=x1y=x-1. Since t+1>1t+1>1, 0<y<20<y<2 and therefore 1<x<31<x<3. The unrestricted line would contain points that the parameter never reaches.

For x=3+2costx=3+2\cos t and y=1+4sinty=-1+4\sin t, cost=x32,sint=y+14.\cos t=\frac{x-3}{2},\qquad \sin t=\frac{y+1}{4}. Therefore (x3)24+(y+1)216=1.\frac{(x-3)^2}{4}+\frac{(y+1)^2}{16}=1. The allowed interval for tt decides whether this represents the whole ellipse or only part of it.

Cartesian form One useful parametrisation
y=F(x)y=F(x) x=t, y=F(t)x=t,\ y=F(t)
(xh)2a2+(yk)2b2=1\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 x=h+acost, y=k+bsintx=h+a\cos t,\ y=k+b\sin t

Eliminating the parameter can enlarge the locus, so a Cartesian equation without the inherited range may be incomplete. Different parameter values can produce the same point, and different parametrisations can describe the same curve. For an intersection, substitute both parametric expressions into the other curve, solve for permitted values of tt, then recover both coordinates.