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Parametric equations questions
Instead of relating y to x directly, both can be given in terms of a third variable, so a curve is traced out as that parameter increases. Parametric equations describe circles, parabolas and flight paths naturally, and eliminating the parameter converts them back to Cartesian form when a question calls for it.
7 original questions · 23 marks · the parametric equations notes · Coordinate geometry
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
A curve has parametric equations x = t2 + 1, y = 2t − 1. Find the coordinates of the point where t = 2.
Worked answer
x = 4 + 1 = 5 and y = 4 − 1 = 3, so the point is (5, 3). One value of the parameter gives one point of the curve. Both coordinates come from the same t, so a mark goes to each substitution. B1 B1, one for each coordinate.Find the Cartesian equation of the curve x = t − 3, y = t2.
Worked answer
The simpler equation gives t = x + 3, and substituting into the other gives y = (x + 3)2. M1 for t = x + 3 substituted, A1 for the Cartesian equation. Solve the easier equation for t and feed the harder one; that route handles almost every algebraic pair. The answer must be free of t, so any t left standing means the conversion is unfinished, however tidy the algebra.A curve has parametric equations x = 2t, y = 12/t for t ≠ 0. Find its Cartesian equation, and name the type of curve.
Worked answer
t = x/2, so y = 12 ÷ (x/2) = 24/x, which rearranges to xy = 24. That is a hyperbola with the axes as asymptotes. M1 for eliminating t, A1 for xy = 24, B1 for naming the hyperbola. The excluded value t = 0 lines up with the curve never reaching x = 0, so nothing extra needs removing. Dividing by a fraction means multiplying by its reciprocal, and answering y = 6/x is the slip that follows from forgetting it.The curve C has parametric equations x = 5 cos t, y = 5 sin t for 0 ≤ t ≤ π. Find the Cartesian equation of C, and state precisely which part of that Cartesian curve C is.
Worked answer
Squaring and adding: x2 + y2 = 25(cos2 t + sin2 t) = 25, the circle of radius 5 about the origin. On 0 ≤ t ≤ π the value y = 5 sin t is never negative, so C is the upper semicircle only, running from (5, 0) at t = 0 to (−5, 0) at t = π. M1 for squaring and adding, A1 for the circle, B1 for the upper semicircle, B1 for the two endpoints. Trig pairs convert by identity, and the parameter's range then says how much of the curve is really there.Find the Cartesian equation of the curve x = 4 cos t, y = 3 sin t.
Worked answer
cos t = x/4 and sin t = y/3, so the identity sin2 t + cos2 t = 1 gives x2/16 + y2/9 = 1. That is an ellipse, 4 wide and 3 tall measured from the centre. M1 for cos t = x/4 and sin t = y/3, dM1 for substituting into the identity, A1 for the ellipse. Divide by the coefficients before squaring; squaring first buries them and the identity no longer fits. Note the denominators are 16 and 9, the squares of 4 and 3.The curve C has parametric equations x = t2, y = t6. Find its Cartesian equation, and explain why C is only part of the curve that equation describes.
Worked answer
y = (t2)3, so y = x3. But x = t2 ≥ 0 for every real t, so C is only the half of the cubic with x ≥ 0. M1 for cubing x = t2, A1 for y = x3, B1 for the restriction x ≥ 0. The Cartesian equation says where a point may be; the parametrisation says where the point actually goes.A ball's flight is modelled by x = 20t, y = 15t − 5t2, with x and y in metres and t in seconds from launch. Find when the ball lands, how far away it lands, and the greatest height it reaches.
Worked answer
Landing means y = 0, and 15t − 5t2 = 5t(3 − t) gives t = 0 at launch or t = 3. So the ball lands after 3 seconds, at x = 20 × 3 = 60 metres. The greatest height sits at the vertex of the y parabola, halfway between the two roots at t = 1.5, where y = 22.5 − 11.25 = 11.25 metres. Differentiating y with respect to t and setting dy/dt = 15 − 10t = 0 reaches the same time. M1 for setting y = 0, A1 for t = 3, B1 for 60 metres, M1 for the vertex of the parabola, A1 for t = 1.5, A1 for 11.25 metres. One clock drives both coordinates, so solve the vertical story for time and then hand that time to the horizontal one. Discarding t = 0 without saying it is the launch, or quoting the greatest height as an x value, are the two ways this loses marks.
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