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Parametric equationsEdexcel A-Level Maths: Revision notes

Section 1

What parametric equations are

In parametric equations the coordinates xx and yy are each given in terms of a third variable, the parameter, usually tt or θ\theta. For each value of the parameter you get one point (x,y)(x,y), and as the parameter changes the point traces out a curve. Example: x=2t+1x=2t+1, y=4t2−3y=4t^2-3. When t=2t=2, (x,y)=(5,13)(x,y)=(5,13). To find where the curve meets an axis, set x=0x=0 (the yy-axis) or y=0y=0 (the xx-axis), solve for tt, then substitute back to find the other coordinate. To find where it meets a line, substitute both parametric equations into the line's equation and solve for tt.

Key termsparameterparametric equations
Common mistake

Finding tt and stopping. Always substitute back to give the coordinates the question asks for.

Section 2

Converting to Cartesian form: eliminating the parameter

To get a Cartesian equation (one linking xx and yy only), eliminate the parameter.

  • Substitution: rearrange one equation for tt and substitute into the other. From x=2t+1x=2t+1, t=x−12t=\frac{x-1}{2}, so y=4(x−12)2−3=x2−2x−2y=4\left(\frac{x-1}{2}\right)^2-3=x^2-2x-2.
  • Multiplying or dividing: for x=5tx=5t, y=5ty=\frac5t, the product xy=25xy=25 removes tt directly.
  • Trigonometric identity: if xx and yy involve cos⁡t\cos t and sin⁡t\sin t, rearrange to get cos⁡t\cos t and sin⁡t\sin t alone and use cos⁡2t+sin⁡2t=1\cos^2t+\sin^2t=1. Useful forms: x=5tx=5t, y=3t2y=3t^2 gives 25y=3x225y=3x^2 (a parabola); x=5tx=5t, y=5ty=\frac5t gives xy=25xy=25 (a hyperbola).
Key termsCartesian equationeliminate
Exam tip

Check the Cartesian equation by substituting one value of tt into both forms.

Common mistake

Squaring and adding when the equations are not of the form acos⁡ta\cos t and asin⁡ta\sin t. First isolate cos⁡t\cos t and sin⁡t\sin t.

Section 3

Circles in parametric form

The equations x=rcos⁡tx=r\cos t, y=rsin⁡ty=r\sin t give the circle x2+y2=r2x^2+y^2=r^2, since r2cos⁡2t+r2sin⁡2t=r2r^2\cos^2t+r^2\sin^2t=r^2. For example x=3cos⁡tx=3\cos t, y=3sin⁡ty=3\sin t is a circle of radius 33 about the origin. A shifted circle is x=a+rcos⁡tx=a+r\cos t, y=b+rsin⁡ty=b+r\sin t, with centre (a,b)(a,b) and radius rr: cos⁡t=x−ar, sin⁡t=y−br ⇒ (x−a)2+(y−b)2=r2.\cos t=\frac{x-a}{r},\ \sin t=\frac{y-b}{r}\ \Rightarrow\ (x-a)^2+(y-b)^2=r^2. The examples x=2+5cos⁡tx=2+5\cos t, y=−4+5sin⁡ty=-4+5\sin t give centre (2,−4)(2,-4) and radius 55. The point starts at the right-hand end of the horizontal diameter when t=0t=0 and moves anticlockwise as tt increases.

Key termscentreradius
Common mistake

Reading the centre with the wrong sign. In x=−2+4cos⁡tx=-2+4\cos t the centre has xx-coordinate −2-2, not 22.

Section 4

The domain of the parameter

The values allowed for tt decide which part of the curve is drawn, so always read the domain.

  • 0≤t<2π0\le t<2\pi gives a full circle, once round.
  • 0≤t≤π0\le t\le\pi for x=2+4cos⁡tx=2+4\cos t, y=−1+4sin⁡ty=-1+4\sin t gives only y≥−1y\ge-1, the upper semicircle.
  • For x=12tx=\frac{12}{t}-type curves, t≠0t\ne0 means the Cartesian form has the restriction x≠0x\ne0.
  • A restricted domain such as −2≤t≤3-2\le t\le3 on x=4tx=4t, y=3t2y=3t^2 gives only part of the parabola 16y=3x216y=3x^2, with −8≤x≤12-8\le x\le12. When you solve for tt, reject any value outside the allowed range. Trigonometric equations usually give two values in [0,2π)[0,2\pi): for cos⁡t=12\cos t=\frac12, t=π3t=\frac{\pi}{3} and 5π3\frac{5\pi}{3}.
Key termsdomainrestricted domain
Common mistake

Giving the Cartesian equation but forgetting the restriction on xx or yy that the parameter's domain creates.

Exam tip

Work out the start point and end point of the curve using the ends of the domain.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Parametric equations

  1. A curve CC has parametric equations x=2t+1x=2t+1, y=4t2−3y=4t^2-3, where tt is a real number.
    Find the coordinates of the point where CC crosses the yy-axis.2 marks
  2. A curve DD has parametric equations x=−2+4cos⁡tx=-2+4\cos t, y=3+4sin⁡ty=3+4\sin t, for 0≤t<2π0\le t<2\pi.
    Find the values of tt, for 0≤t<2π0\le t<2\pi, at the points where DD meets the line x=0x=0.2 marks
  3. A curve HH has parametric equations x=3tx=3t, y=12ty=\frac{12}{t}, where t≠0t\ne0.
    Find a Cartesian equation of HH in the form xy=kxy=k, and state the values that xx cannot take.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).