Questions on parametric equations

Question 1

A curve has parametric equations x = t² + 1, y = 2t − 1. Find the coordinates of the point where t = 2.

Accepted Answer

x = 4 + 1 = 5 and y = 4 − 1 = 3, so the point is (5, 3). B1 B1, one for each coordinate.

Question 2

Find the Cartesian equation of the curve x = t − 3, y = t².

Accepted 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.

Question 3

A curve has parametric equations x = t² − 4, y = 2t − 6. Find the value of t for which the curve crosses the x-axis, and state the coordinates of that point.

Accepted Answer

The x-axis has y = 0, so 2t − 6 = 0 gives t = 3. M1 for setting y = 0, A1 for t = 3. Substituting into x, x = 3² − 4 = 5, so the curve crosses the x-axis at (5, 0). A1 for the point.

Question 4

A curve has parametric equations x = 4 − t, y = t² for −2 ≤ t ≤ 2. Find the coordinates of the points where t = −2, 0 and 2, and state whether x increases or decreases as t increases.

Accepted Answer

At t = −2: x = 6, y = 4, giving (6, 4). At t = 0: x = 4, y = 0, giving (4, 0). At t = 2: x = 2, y = 4, giving (2, 4). M1 for substituting all three values of t, A1 for the three points. Since x = 4 − t, x decreases as t increases, so the curve is traced from right to left as t runs from −2 to 2. B1 for the direction.

Question 5

A curve has parametric equations x = 2t, y = 12/t for t ≠ 0. Find its Cartesian equation, and name the type of curve.

Accepted 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.

Question 6

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.

Accepted 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.

Question 7

Find the Cartesian equation of the curve x = 4 cos t, y = 3 sin t.

Accepted 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.

Question 8

A curve has parametric equations x = 2 sec θ, y = 3 tan θ. Using the identity 1 + tan²θ = sec²θ, find the Cartesian equation of the curve, and name the type of curve it is.

Accepted Answer

sec θ = x/2 and tan θ = y/3, so 1 + (y/3)² = (x/2)², that is x²/4 − y²/9 = 1. M1 for sec θ = x/2 and tan θ = y/3, dM1 for substituting into the identity, A1 for the equation. This is a hyperbola. B1 for naming it.

Question 9

A curve has parametric equations x = 1/t, y = t² for t ≠ 0. Find its Cartesian equation, and state the values of x for which points of the curve exist.

Accepted Answer

t = 1/x, so y = (1/x)² = 1/x². M1 for t = 1/x substituted, A1 for y = 1/x². Since t ranges over every nonzero real number, x = 1/t also ranges over every nonzero real number, so the curve exists for all x ≠ 0, the whole of the Cartesian curve rather than half of it. B1 for the domain, correctly stated as no extra restriction beyond x ≠ 0.

Question 10

The curve C has parametric equations x = 4 cos θ, y = 4 sin θ for 0 ≤ θ < 2π. Find the values of θ, and the coordinates of the points, where C meets the line y = x.

Accepted Answer

Setting 4 sin θ = 4 cos θ gives tan θ = 1, so θ = π/4 or θ = 5π/4 in the given range. M1 for tan θ = 1, A1 for both angles. At θ = π/4, x = 4 cos(π/4) = 2√2 and y = 2√2, giving (2√2, 2√2). At θ = 5π/4, x = y = −2√2, giving (−2√2, −2√2). M1 for substituting back, A1 for both points. The two points are diametrically opposite, since π/4 and 5π/4 differ by π, half a full turn around the circle.

Question 11

A particle's position at time t seconds is given by x = t² − 2t, y = 3t, with x and y in metres and t ≥ 0. Find the particle's position at t = 4, and the other time at which x = 0.

Accepted Answer

At t = 4, x = 16 − 8 = 8 and y = 12, giving (8, 12). M1 for substituting t = 4, A1 for the point. Setting x = 0 gives t² − 2t = 0, that is t(t − 2) = 0, so t = 0 or t = 2. The particle starts at x = 0 when t = 0, so the other time is t = 2 seconds, at which y = 6, so it returns to the y-axis at (0, 6). M1 for solving x = 0, A1 for t = 2 with the point.

Question 12

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.

Accepted 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.

Question 13

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.

Accepted 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.

Question 14

A curve has parametric equations x = cos 2t, y = sin t. Using the identity cos 2t = 1 − 2 sin²t, find the Cartesian equation of the curve, state the values of y for which the curve exists, and hence find the two values of y at which x = 0.5.

Accepted Answer

cos 2t = 1 − 2 sin²t, and sin t = y, so x = 1 − 2y². M1 for the identity, A1 for x = 1 − 2y². Since sin t takes every value in [−1, 1] as t varies, and no value outside that range is possible, the curve exists for −1 ≤ y ≤ 1, the whole of that parabola arc rather than part of it. M1 for the range of sin t, A1 for the domain with the reasoning. Setting x = 0.5: 1 − 2y² = 0.5, so y² = 0.25 and y = 0.5 or y = −0.5, both inside [−1, 1] and so both genuinely on the curve. M1 for solving for y, A1 for both values.

Question 15

A curve has parametric equations x = t + 1/t, y = t − 1/t for t ≠ 0. Show that its Cartesian equation is x² − y² = 4, and find the exact value(s) of y when x = 5.

Accepted Answer

x² − y² = (t + 1/t)² − (t − 1/t)² = (t² + 2 + 1/t²) − (t² − 2 + 1/t²) = 4. M1 for expanding both squares, M1 for the subtraction, A1 for x² − y² = 4. Substituting x = 5: 25 − y² = 4, so y² = 21. M1 for the substitution, A1 for y² = 21, and y = ±√21. A1 for both values.