Edexcel International A Level Maths · P4 · Pure Mathematics 4
Parametric equations
28 free, high-quality past-paper-style questions on Parametric equations for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 271 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P4-2-B-01P4 · A2Easy6 marks
A curve C has parametric equations x = 3t + 2, y = t2 − 1, t ∈ ℝ.
(a)
Find the coordinates of the point on C where t = 2.
(1)
(b)
Find a Cartesian equation of C in the form y = ax2 + bx + c, where a, b and c are constants.
(3)
(c)
Find the x-coordinates of the points where C meets the x-axis.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
(8, 3)
B1
Part (b)
t = (x − 2)/3
M1
substitutes: y = ((x − 2)/3)2 − 1
M1
y = (1/9)x2 − (4/9)x − 5/9
A1
Part (c)
y = 0 ⇒ t2 = 1 ⇒ t = ±1, or solves x2 − 4x − 5 = 0
M1
x = 5 and x = −1
A1
Model sketch
Figure: the curve C: x = 3t + 2, y = t² − 1
Exam tip: Make the parameter the subject of the linear equation, never the quadratic one: it avoids a ± and keeps the algebra short.
Figure 1 shows the curve C with parametric equations x = 2t + a, y = b − t2, 0 ≤ t ≤ 3, where a and b are constants. The end points of C are A, where t = 0, and B, where t = 3. Both lie on gridline intersections.
(a)
Use Figure 1 to find the value of a and the value of b.
(2)
(b)
Find a Cartesian equation of C in the form y = px2 + qx + r, where p, q and r are constants.
(3)
(c)
Write down the set of values of x and the set of values of y on C.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
A is (−1, 4) and at t = 0, x = a, so a = −1
B1
y = b at t = 0, so b = 4 (B at t = 3 gives (5, −5), consistent)
B1
Part (b)
t = (x + 1)/2
M1
substitutes: y = 4 − ((x + 1)/2)2
M1
y = −¼x2 − ½x + 15/4
A1
Part (c)
−1 ≤ x ≤ 5
B1
y decreases from 4 to −5 as t increases from 0 to 3, so −5 ≤ y ≤ 4
B1
Exam tip: The ends of a parametric curve come from the ends of the parameter interval. Substitute t = 0 and t = 3, then check whether x or y turns back in between before writing the set of values.
Figure 1 shows the circle C with parametric equations x = a + r cos θ, y = b + r sin θ, 0 ≤ θ < 2π, where a, b and r are positive constants. K is the centre of C. The point P lies on C at a gridline intersection.
(a)
Use Figure 1 to write down the values of a, b and r.
(2)
(b)
By eliminating θ, obtain a Cartesian equation of C.
(2)
(c)
Find the value of θ at P, giving your answer in radians to 3 significant figures.
(2)
(d)
Find dy/dx in terms of θ, and hence find the gradient of the tangent to C at P.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
K is (1, 2), so a = 1, b = 2
B1
C reaches from x = −4 to x = 6, so r = 5
B1
Part (b)
cos θ = (x − 1)/5, sin θ = (y − 2)/5 and uses sin2θ + cos2θ = 1
M1
(x − 1)2 + (y − 2)2 = 25
A1
Part (c)
P is (4, 6): cos θ = 3/5 and sin θ = 4/5, both positive
M1
θ = 0.927
A1
Part (d)
dy/dx = (5 cos θ)/(−5 sin θ) = −cot θ
M1
at P: −(3/5)/(4/5) = −3/4
A1
Exam tip: To find θ at a point, use both cos θ and sin θ. One equation alone gives two angles in [0, 2π); only one of them has the right signs for both.
Figure 1 shows the curve C with parametric equations x = 1 + t2, y = 4t − t2, 0 ≤ t ≤ 3. The point P lies on C at a gridline intersection, and M is the highest point of C.
(a)
Use Figure 1 to find the value of t at P.
(2)
(b)
Find dy/dx in terms of t.
(2)
(c)
Find an equation of the tangent to C at P.
(2)
(d)
Find the coordinates of M.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
P is (2, 3), so 1 + t2 = 2 ⇒ t = 1 (since t ≥ 0)
M1
check: y = 4 − 1 = 3, so t = 1
A1
Part (b)
dx/dt = 2t and dy/dt = 4 − 2t
M1
dy/dx = (4 − 2t)/(2t) = (2 − t)/t
A1
Part (c)
gradient at P = (2 − 1)/1 = 1
M1
y − 3 = 1(x − 2), so y = x + 1
A1
Part (d)
dy/dt = 0 ⇒ t = 2 (and dx/dt ≠ 0)
M1
M is (5, 4)
A1
Common error: Setting dx/dt = 0 to find the highest point. A tangent parallel to the x-axis needs dy/dt = 0 with dx/dt ≠ 0.
Figure 1 shows the curve C with parametric equations x = t2, y = t + 1, t ≥ 0. The shaded region R is bounded by C, the x-axis and the lines x = 1 and x = 4. These lines meet C at P and Q, and the chord PQ is shown dashed.
(a)
Show that the area of R is given by ∫12 (2t2 + 2t) dt.
(3)
(b)
Hence find the exact area of R.
(2)
(c)
Find the area of the region between C and the chord PQ.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
dx/dt = 2t
B1
area = ∫ y (dx/dt) dt = ∫ (t + 1)(2t) dt = ∫ (2t2 + 2t) dt
M1
x = 1 ⇒ t = 1 and x = 4 ⇒ t = 2 (as t ≥ 0), giving the limits, as required
A1
Part (b)
[2t3/3 + t2]12
M1
= (16/3 + 4) − (2/3 + 1) = 23/3
A1
Part (c)
P is (1, 2) and Q is (4, 3)
B1
area under PQ (a trapezium) = ½(2 + 3) × 3 = 15/2
M1
23/3 − 15/2 = 1/6
A1
Exam tip: Change the x-limits into t-limits before you integrate. An integral with respect to t needs t-limits.
A curve C has parametric equations x = 3 cos t, y = 2 sin 2t, 0 ≤ t ≤ π/2.
(a)
Show that a Cartesian equation of C is y2 = (16/81)x2(9 − x2).
(3)
(b)
Find the set of values of x and the set of values of y on C.
(3)
(c)
Find the exact coordinates of the point on C at which the tangent is parallel to the x-axis.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
y = 4 sin t cos t, using sin 2t = 2 sin t cos t
M1
y2 = 16 sin2t cos2t = 16(1 − cos2t) cos2t
M1
cos t = x/3 ⇒ y2 = 16(1 − x2/9)(x2/9) = (16/81)x2(9 − x2) as required
A1
Part (b)
cos t decreases from 1 to 0, so 0 ≤ x ≤ 3
B1
2t runs from 0 to π, so sin 2t rises from 0 to 1 (at t = π/4) and returns to 0
M1
0 ≤ y ≤ 2
A1
Part (c)
dy/dx = 4 cos 2t/(−3 sin t)
M1
cos 2t = 0 ⇒ t = π/4
M1
(3√2/2, 2)
A1
Model sketch
Figure: C: x = 3 cos t, y = 2 sin 2t for 0 ≤ t ≤ π/2
Exam tip: When the y-equation has sin 2t and the x-equation has cos t, square y so that only sin2 t and cos2 t remain, then replace sin2 t by 1 − cos2 t.
A curve C has parametric equations x = sec t, y = sin t, 0 ≤ t ≤ π/3. C meets the x-axis at (1, 0). The region R is bounded by C, the x-axis and the line x = 2.
Figure: the region R
(a)
Find the value of t at the point of C where x = 2.
(1)
(b)
Show that the area of R is given by ∫0π/3 tan2t dt.
(3)
(c)
Hence find the exact area of R.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
sec t = 2 ⇒ cos t = ½ ⇒ t = π/3
B1
Part (b)
dx/dt = sec t tan t
B1
y (dx/dt) = sin t × (1/cos t) × tan t = tan t × tan t = tan2t
M1
x = 1 at t = 0 and x = 2 at t = π/3 give the limits, as required
A1
Part (c)
tan2t = sec2t − 1
M1
∫ (sec2t − 1) dt = [tan t − t]
M1
(√3 − π/3) − 0 = √3 − π/3
A1
Exam tip: tan2 t cannot be integrated directly. Replace it by sec2 t − 1, whose integral is tan t − t.
For each value of the parameter t ≥ 0, a point P has coordinates given by x = (1 − t2)/(1 + t2) and y = 2t/(1 + t2). As t increases from 0, P traces out a curve C.
(a)
Show that every point of C lies on the circle x2 + y2 = 1.
(3)
(b)
Explain why no value of t gives the point (−1, 0).
(2)
(c)
Find the set of values of x and the set of values of y on C.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
x2 + y2 = [(1 − t2)2 + 4t2]/(1 + t2)2
M1
numerator = 1 + 2t2 + t4 = (1 + t2)2
M1
so x2 + y2 = 1 as required
A1
Part (b)
x = −1 would need 1 − t2 = −1 − t2, i.e. 1 = −1
M1
which is impossible, so (−1, 0) is not on C
A1
Part (c)
x = 2/(1 + t2) − 1 decreases from 1 towards −1 without reaching it: −1 < x ≤ 1
B1
y ≥ 0 for t ≥ 0, and 1 − y = (1 − t)2/(1 + t2) ≥ 0 so y ≤ 1, with y = 1 at t = 1
M1
0 ≤ y ≤ 1
A1
Model sketch
Figure: C is the upper half of the unit circle, starting at (1, 0) and never reaching (−1, 0)
Exam tip: Writing x = 2/(1 + t2) − 1 makes its behaviour obvious: as t grows, the fraction falls towards 0, so x falls towards −1 but never reaches it.
Figure 1 shows the curve C with parametric equations x = sin t + cos t, y = sin 2t, 0 ≤ t ≤ π. Starting at A when t = 0, the curve rises to V, turns back through A, passes through Q and ends at E when t = π. The point Q lies on a gridline intersection.
(a)
Show that a Cartesian equation of C is y = x2 − 1.
(3)
(b)
By writing x in the form R sin(t + α), find the set of values of x on C.
(3)
(c)
Use Figure 1 to write down the coordinates of Q, and find the value of t at Q.
(2)
(d)
Show that dy/dx = 2(sin t + cos t) for t ≠ π/4, and explain how this agrees with part (a).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
x2 = sin2t + 2 sin t cos t + cos2t
M1
uses sin2t + cos2t = 1 and 2 sin t cos t = sin 2t: x2 = 1 + sin 2t
M1
so y = sin 2t = x2 − 1, as required
A1
Part (b)
x = √2 sin(t + π/4)
B1
t + π/4 runs from π/4 to 5π/4, so sin(t + π/4) rises from √2/2 to 1 and falls to −√2/2
M1
−1 ≤ x ≤ √2
A1
Part (c)
Q is (0, −1), so sin 2t = −1 with 0 ≤ 2t ≤ 2π ⇒ 2t = 3π/2
M1
t = 3π/4 (and x = √2/2 − √2/2 = 0 ✓)
A1
Part (d)
dy/dx = 2 cos 2t/(cos t − sin t) = 2(cos t − sin t)(cos t + sin t)/(cos t − sin t)
M1
= 2(sin t + cos t) = 2x, which is the gradient of y = x2 − 1
A1
Exam tip: When both x and y involve sin t and cos t, try squaring x: the cross term 2 sin t cos t is sin 2t.
Figure 1 shows the curve C with parametric equations x = 2t, y = 3/t, t ≠ 0. The point P on C lies at a gridline intersection in the first quadrant. The normal to C at P meets C again at the point Q.
(a)
Use Figure 1 to find the value of t at P.
(1)
(b)
Find dy/dx in terms of t.
(2)
(c)
Show that the normal to C at P has equation 2x − 3y + 5 = 0.
Exam tip: To find where a line meets a parametric curve, substitute x(t) and y(t) into the line. You already know one root (the parameter at P), so the quadratic factorises.
Figure 1 shows the curve C with parametric equations x = 2 + 1/t, y = 3 − 2t, t > 0, and the dashed lines x = 2 and y = 3. The straight line L passes through the two marked points, which lie on gridline intersections.
(a)
Find a Cartesian equation of C in the form y = (ax + b)/(x − 2), where a and b are integers.
(3)
(b)
Find the set of values of x and the set of values of y on C.
(2)
(c)
Use Figure 1 to find an equation of L.
(1)
(d)
Find the exact coordinates of the points where L meets C.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
t = 1/(x − 2)
M1
y = 3 − 2/(x − 2)
M1
y = (3x − 8)/(x − 2)
A1
Part (b)
1/t takes all positive values, so x > 2
B1
2t takes all positive values, so y < 3
B1
Part (c)
L passes through (3, 0) and (5, 2), so y = x − 3
B1
Part (d)
3 − 2t = (2 + 1/t) − 3 ⇒ 2t2 − 4t + 1 = 0
M1
t = (2 ± √2)/2, both positive
A1
(4 − √2, 1 − √2) and (4 + √2, 1 + √2)
A1
Common error: Ignoring the condition t > 0. It is what makes C a single branch with x > 2 and y < 3, and each root of the quadratic must be checked against it.
Figure 1 shows the curve C with parametric equations x = tan t, y = cos2t, −π/2 < t < π/2. The shaded region R is bounded by C, the x-axis and the lines x = −1 and x = √3.
(a)
Show that a Cartesian equation of C is y = 1/(1 + x2).
(2)
(b)
Show that the area of R is equal to ∫ 1 dt between suitable limits, and hence find the exact area of R.
(4)
(c)
Find dy/dx in terms of t, and hence find the gradient of C at the point where x = 1.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
y = cos2t = 1/sec2t and sec2t = 1 + tan2t
M1
so y = 1/(1 + x2), as required
A1
Part (b)
dx/dt = sec2t
B1
y (dx/dt) = cos2t sec2t = 1
M1
x = −1 ⇒ t = −π/4 and x = √3 ⇒ t = π/3
B1
area = π/3 − (−π/4) = 7π/12
A1
Part (c)
dy/dt = −2 cos t sin t
M1
dy/dx = −2 sin t cos t ÷ sec2t = −2 sin t cos3t
A1
x = 1 ⇒ t = π/4: −2(√2/2)(√2/2)3 = −1/2
A1
Exam tip: The parametric form can make an area easy when the Cartesian form is hard: here y × dx/dt simplifies to 1.
A drone takes off from the point O on level ground and flies in a vertical plane. Its path is modelled by the parametric equations x = 8t, y = 12t − 2t2, 0 ≤ t ≤ 6, where x m and y m are the horizontal and vertical distances from O and t is the time in seconds. Figure 1 shows the path and a vertical wall standing on the ground. The top of the wall, W, lies on a gridline intersection.
(a)
Find a Cartesian equation of the path in the form y = px − qx2, where p and q are constants.
(2)
(b)
Find the greatest height of the drone above the ground, and the value of x at which it occurs.
(3)
(c)
Use Figure 1 to find how far the drone passes above the top of the wall.
(2)
(d)
Find the angle between the path of the drone and the horizontal as it passes over the wall.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
t = x/8
M1
y = (3/2)x − x2/32
A1
Part (b)
dy/dx = (12 − 4t)/8
M1
dy/dx = 0 ⇒ t = 3
A1
greatest height 18 m, at x = 24
A1
Part (c)
W is (40, 8); x = 40 ⇒ t = 5
M1
y = 60 − 50 = 10, so the drone passes 2 m above W
A1
Part (d)
at t = 5, dy/dx = (12 − 20)/8 = −1
M1
45° (the drone is descending)
A1
Exam tip: The gradient dy/dx gives the direction of motion along a path. The angle with the horizontal is arctan |dy/dx|; its sign tells you whether the path is rising or falling.
Common error: Counting θ = 0 and θ = π as turning points. There sin θ = 0 as well, so the formula gives 0/0: these are the end points of C, not turning points.
A curve C has parametric equations x = 1 − t2, y = t − t3, −1 ≤ t ≤ 1. C is a closed loop that passes through the origin O when t = −1 and when t = 1, and lies in the region 0 ≤ x ≤ 1.
(a)
Show that a Cartesian equation of C is y2 = x2(1 − x).
(3)
(b)
Find the exact coordinates of the points on C at which the tangent is parallel to the x-axis.
(4)
(c)
Find the equations of the two tangents to C at O.
(2)
(d)
Find the exact area of the region enclosed by the loop.
(5)
Show mark scheme
Answer
Mark
Notes
Part (a)
y = t(1 − t2) = tx
M1
y2 = t2x2
M1
t2 = 1 − x ⇒ y2 = x2(1 − x) as required
A1
Part (b)
dx/dt = −2t and dy/dt = 1 − 3t2
B1
dy/dx = (1 − 3t2)/(−2t)
M1
1 − 3t2 = 0 ⇒ t = ±1/√3
M1
(2/3, 2√3/9) and (2/3, −2√3/9)
A1
Part (c)
at t = 1 the gradient is (1 − 3)/(−2) = 1; at t = −1 it is (1 − 3)/2 = −1
M1
y = x and y = −x
A1
Part (d)
dx/dt = −2t
B1
upper half (y ≥ 0) is 0 ≤ t ≤ 1, with x = 0 at t = 1 and x = 1 at t = 0: area = ∫10 (t − t3)(−2t) dt = ∫01 (2t2 − 2t4) dt
M1
= 2/3 − 2/5 = 4/15
A1
the loop is symmetric in the x-axis, so doubles
M1
area = 8/15
A1
Model sketch
Figure: the loop with its horizontal tangents and the two tangents at O
Exam tip: For a loop, find the area of one half and double it. Choose the half with y ≥ 0, and set the limits from the left-hand end (x = 0) to the right-hand end (x = 1) in terms of t.
A curve C has parametric equations x = 2t − sin 2t, y = 1 − cos 2t, 0 ≤ t ≤ π. C forms a single arch, starting at the origin and meeting the x-axis again at (2π, 0).
Figure: the arch C and the region between C and the x-axis
(a)
Show that dy/dx = cot t, for 0 < t < π.
(4)
(b)
Find an equation of the tangent to C at the point where t = π/4, giving your answer in the form y = x + k, where k is an exact constant.
(3)
(c)
Find the exact area of the region bounded by C and the x-axis.
(6)
Show mark scheme
Answer
Mark
Notes
Part (a)
dx/dt = 2 − 2 cos 2t and dy/dt = 2 sin 2t
B1
dy/dx = 2 sin 2t/(2 − 2 cos 2t)
M1
uses sin 2t = 2 sin t cos t and 1 − cos 2t = 2 sin2t
M1
= (2 sin t cos t)/(2 sin2t) = cot t as required
A1
Part (b)
point (π/2 − 1, 1)
B1
gradient cot(π/4) = 1 and y − 1 = x − (π/2 − 1)
M1
y = x + 2 − π/2
A1
Part (c)
area = ∫0π (1 − cos 2t)(2 − 2 cos 2t) dt = 2∫0π (1 − cos 2t)2 dt
M1
expands: 2∫ (1 − 2 cos 2t + cos2 2t) dt
M1
uses cos2 2t = ½(1 + cos 4t)
M1
2[(3/2)t − sin 2t + (1/8) sin 4t]
A1
substitutes the limits π and 0
M1
= 3π
A1
Exam tip: A square of a cosine cannot be integrated directly: use cos2 A = ½(1 + cos 2A) first. Here A = 2t, so the new term is cos 4t.
A curve C has parametric equations x = t2, y = t3, t ∈ ℝ. The point P on C has parameter t = 2.
(a)
Find a Cartesian equation of C in the form y2 = f(x).
(2)
(b)
Find dy/dx in terms of t.
(2)
(c)
Find an equation of the tangent to C at P.
(3)
(d)
The tangent at P meets C again at the point Q. Show that the parameter of Q satisfies t3 − 3t2 + 4 = 0, and hence find the coordinates of Q.
(4)
(e)
Find the area of the finite region bounded by C, the x-axis and the line x = 4, for which y ≥ 0.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
y2 = t6 = (t2)3
M1
y2 = x3
A1
Part (b)
3t2/(2t)
M1
dy/dx = 3t/2 (t ≠ 0)
A1
Part (c)
P(4, 8)
B1
gradient 3 and y − 8 = 3(x − 4)
M1
y = 3x − 4
A1
Part (d)
substitutes x = t2, y = t3 into the tangent: t3 = 3t2 − 4
M1
t3 − 3t2 + 4 = 0 as required
A1
t = 2 is a repeated root (tangent at P): (t − 2)2(t + 1) = 0
M1
t = −1, Q = (1, −1)
A1
Part (e)
∫02t3 × 2t dt = ∫02 2t4 dt
M1
[2t5/5]02
M1
= 64/5
A1
Model sketch
Figure: C: y² = x³, the tangent at P meeting C again at Q(1, −1), and the region for part (e)
Exam tip: When a tangent at the point with parameter p meets the curve again, the equation in t always has p as a repeated root. Divide out (t − p)2 to find the other root quickly.
For positive values of t, the equations x = ln t and y = t + 1/t define a curve C, which lies entirely above the x-axis.
(a)
Find a Cartesian equation of C in the form y = f(x), and state the domain of f.
(3)
(b)
Show that y ≥ 2 for every point on C, and state the coordinates of the point on C where y = 2.
(3)
(c)
Find an equation of the normal to C at the point where t = 2, in the form y = ax + b, where a and b are exact constants.
(4)
(d)
Find the exact area of the region bounded by C, the x-axis, the y-axis and the line x = ln 3.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
t = ex
M1
y = ex + e−x
A1
ln t takes every real value for t > 0, so the domain is x ∈ ℝ
B1
Part (b)
y − 2 = t + 1/t − 2 = (t − 1)2/t
M1
(t − 1)2 ≥ 0 and t > 0, so y − 2 ≥ 0, i.e. y ≥ 2
A1
equality when t = 1: (0, 2)
B1
Part (c)
point (ln 2, 5/2)
B1
dy/dx = (1 − t−2)/(1/t) = t − 1/t
M1
at t = 2 the gradient is 3/2, so the normal gradient is −2/3
M1
y = −(2/3)x + 5/2 + (2/3) ln 2
A1
Part (d)
x = 0 ⇒ t = 1 and x = ln 3 ⇒ t = 3
B1
∫13 (t + 1/t)(1/t) dt = ∫13 (1 + t−2) dt
M1
[t − 1/t]13
M1
= (3 − 1/3) − 0 = 8/3
A1
Model sketch
Figure: C: y = eˣ + e⁻ˣ, the normal at P(ln 2, 5/2) where t = 2, and the region for part (d)
Exam tip: To prove an inequality such as t + 1/t ≥ 2, subtract 2 and write the result over a common denominator: the numerator becomes a perfect square.
Figure 1 shows the curve C with parametric equations x = 8 cos3t, y = 8 sin3t, 0 ≤ t ≤ π/2. The point P on C has parameter t = π/3. The tangent to C at P meets the x-axis at A and the y-axis at B. The shaded region R is bounded by C and the coordinate axes.
(a)
Show that dy/dx = −tan t.
(3)
(b)
Find the coordinates of P, and show that the tangent to C at P has equation y = −√3x + 4√3.
(3)
(c)
Find the length AB.
(2)
(d)
Show that the area of R is given by 192 ∫0π/2 sin4t cos2t dt.
(2)
(e)
Given that sin4t cos2t = ⅛(sin2 2t − sin2 2t cos 2t), find the exact area of R.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
differentiates both by the chain rule
M1
dx/dt = −24 cos2t sin t and dy/dt = 24 sin2t cos t
A1
dy/dx = (24 sin2t cos t)/(−24 cos2t sin t) = −sin t/cos t = −tan t, as required
A1
Part (b)
P is (8 × ⅛, 8 × 3√3/8) = (1, 3√3)
B1
gradient = −tan(π/3) = −√3
M1
y − 3√3 = −√3(x − 1) ⇒ y = −√3x + 4√3, as required
A1
Part (c)
A is (4, 0) and B is (0, 4√3)
B1
AB = √(42 + (4√3)2) = √64 = 8
B1
Part (d)
area = ∫ y (dx/dt) dt from t = π/2 (where x = 0) to t = 0 (where x = 8)
M1
= ∫π/20 8 sin3t (−24 cos2t sin t) dt = 192 ∫0π/2 sin4t cos2t dt, as required
∫ sin2 2t cos 2t dt = ⅙ sin3 2t, which is 0 at both limits
M1
area = 192 × ⅛ × (π/4 − 0) = 6π
A1
Common error: Taking the limits 0 to π/2 straight from the parameter interval. At t = 0 the curve is at x = 8, so as x runs from 0 to 8, t runs from π/2 down to 0. Reversing the limits cancels the minus sign in dx/dt.
Figure 1 shows the curve C with parametric equations x = t2 + 2t, y = t2 − 2t, t ∈ ℝ. The tangent to C at V is parallel to the y-axis and the tangent to C at L is parallel to the x-axis. The shaded region R is bounded by C and the x-axis.
(a)
Show that a Cartesian equation of C is (x − y)2 = 8(x + y).
(3)
(b)
Find the coordinates of the points where C meets the x-axis.
(2)
(c)
Find dy/dx in terms of t.
(2)
(d)
Find the coordinates of V and of L.
(3)
(e)
Find the area of R.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
x − y = 4t
M1
x + y = 2t2
M1
t = (x − y)/4 ⇒ x + y = 2(x − y)2/16 ⇒ (x − y)2 = 8(x + y), as required
A1
Part (b)
t2 − 2t = 0 ⇒ t = 0 or t = 2
M1
(0, 0) and (8, 0)
A1
Part (c)
dx/dt = 2t + 2 and dy/dt = 2t − 2
M1
dy/dx = (t − 1)/(t + 1)
A1
Part (d)
V where dx/dt = 0 and L where dy/dt = 0
M1
t = −1: V(−1, 3)
A1
t = 1: L(3, −1)
A1
Part (e)
area from ∫ y (dx/dt) dt with limits t = 0 and t = 2 (x increases from 0 to 8)
M1
(t2 − 2t)(2t + 2) = 2t3 − 2t2 − 4t
A1
[t4/2 − 2t3/3 − 2t2]02 = 8 − 16/3 − 8 = −16/3
M1
R lies below the x-axis, so its area is 16/3
A1
Exam tip: A region below the x-axis gives a negative value of ∫ y (dx/dt) dt. Say why it is negative, and give the area as a positive number.
Figure 1 shows the curve C with parametric equations x = sin t, y = tan t, −π/2 < t < π/2. The dashed lines are asymptotes to C. The point P on C has parameter t = π/3. The shaded region R is bounded by C, the x-axis and the line through P parallel to the y-axis.
(a)
Show that a Cartesian equation of C is y = x/√(1 − x2).
(3)
(b)
Write down the equations of the two asymptotes, and explain why no point of C has an x-coordinate outside them.
(2)
(c)
Show that dy/dx = sec3t, and hence state the least value of the gradient of C.
(3)
(d)
Find an equation of the tangent to C at P, in the form y = mx + c, where m and c are exact.
(2)
(e)
Find the area of R.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
y = sin t/cos t
M1
cos t = +√(1 − sin2t), the positive root because cos t > 0 for −π/2 < t < π/2
M1
y = x/√(1 − x2), as required
A1
Part (b)
x = −1 and x = 1
B1
for −π/2 < t < π/2, −1 < sin t < 1, so −1 < x < 1; as t → ±π/2, x → ±1 and y → ±∞
B1
Part (c)
dx/dt = cos t and dy/dt = sec2t
M1
dy/dx = sec2t/cos t = sec3t, as required
A1
0 < cos t ≤ 1 so sec3t ≥ 1: least gradient 1, at O
A1
Part (d)
P is (√3/2, √3) and the gradient is sec3(π/3) = 23 = 8
M1
y − √3 = 8(x − √3/2) ⇒ y = 8x − 3√3
A1
Part (e)
area = ∫0π/3 tan t cos t dt
M1
= ∫0π/3 sin t dt = [−cos t]0π/3
M1
= −½ + 1 = ½
A1
Common error: Writing cos t = ±√(1 − x2) and keeping both signs. The interval −π/2 < t < π/2 makes cos t positive, which is why C is a single curve.
The opening under a bridge is modelled by the semi-elliptical arch with parametric equations x = 10 cos t, y = 4 sin t, 0 ≤ t ≤ π, where x m and y m are measured from O, the point on the water line directly below the top of the arch. Figure 1 shows the arch and a rectangle of width 12 m and height h m, which models the cross-section of a boat. The rectangle stands on the water line, placed symmetrically about the y-axis, and its upper corners D and E lie on the arch.
(a)
Show that a Cartesian equation of the arch is x2/100 + y2/16 = 1.
(2)
(b)
Find the value of h.
(2)
(c)
Find the gradient of the arch at E.
(3)
(d)
Find the exact area of the opening under the arch, above the water line.
(3)
(e)
Find the area of the part of the opening that lies directly above the rectangle, giving your answer to 3 significant figures.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.