Hard A-Level Parametric curves Questions

Challenging, exam-style A-Level Parametric curves questions with worked solutions. Stretch yourself on the hardest parametric, cartesian, ellipse, hyperbola problems.

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A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
The curve x=5cos⁡(t)+4x=5 \cos{\left(t \right)} + 4, y=5sin⁡(t)−3y=5 \sin{\left(t \right)} - 3 can be written in Cartesian form. Select the correct equation.
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Worked solution

  1. Write \cos t and \sin t in terms of x and y

    cos⁡t=x−45,sin⁡t=y+35\cos t=\frac{x - 4}{5},\quad \sin t=\frac{y + 3}{5}

    Rearrange each parametric equation for the trig ratio.

  2. Apply the identity \cos^{2}t+\sin^{2}t=1

    (x−45)2+(y+35)2=1\left(\frac{x - 4}{5}\right)^{2}+\left(\frac{y + 3}{5}\right)^{2}=1

    Squaring and adding gives the equation of a circle.

  3. Recall the method

    cos⁡2t+sin⁡2t=1\cos^{2}t+\sin^{2}t=1

    The Pythagorean identity gives the circle's equation.

  4. Evaluate the coordinates at t=0

    t=0: (x,y)=(9, −3)t=0:\ (x,y)=\left(9,\ -3\right)

    Substituting the parameter value gives a point on the curve.

  5. Evaluate the coordinates at t=\frac{\pi}{6}

    t=π6: (x,y)=(4+532, −12)t=\frac{\pi}{6}:\ (x,y)=\left(4 + \frac{5 \sqrt{3}}{2},\ - \frac{1}{2}\right)

    Substituting the parameter value gives a point on the curve.

  6. Evaluate the coordinates at t=\frac{\pi}{4}

    t=π4: (x,y)=(522+4, −3+522)t=\frac{\pi}{4}:\ (x,y)=\left(\frac{5 \sqrt{2}}{2} + 4,\ -3 + \frac{5 \sqrt{2}}{2}\right)

    Substituting the parameter value gives a point on the curve.

  7. Evaluate the coordinates at t=\frac{\pi}{3}

    t=π3: (x,y)=(132, −3+532)t=\frac{\pi}{3}:\ (x,y)=\left(\frac{13}{2},\ -3 + \frac{5 \sqrt{3}}{2}\right)

    Substituting the parameter value gives a point on the curve.

  8. Evaluate the coordinates at t=\frac{\pi}{2}

    t=π2: (x,y)=(4, 2)t=\frac{\pi}{2}:\ (x,y)=\left(4,\ 2\right)

    Substituting the parameter value gives a point on the curve.

  9. Evaluate the coordinates at t=\frac{2 \pi}{3}

    t=2π3: (x,y)=(32, −3+532)t=\frac{2 \pi}{3}:\ (x,y)=\left(\frac{3}{2},\ -3 + \frac{5 \sqrt{3}}{2}\right)

    Substituting the parameter value gives a point on the curve.

  10. Evaluate the coordinates at t=\frac{3 \pi}{4}

    t=3π4: (x,y)=(4−522, −3+522)t=\frac{3 \pi}{4}:\ (x,y)=\left(4 - \frac{5 \sqrt{2}}{2},\ -3 + \frac{5 \sqrt{2}}{2}\right)

    Substituting the parameter value gives a point on the curve.

  11. Evaluate the coordinates at t=\frac{5 \pi}{6}

    t=5π6: (x,y)=(4−532, −12)t=\frac{5 \pi}{6}:\ (x,y)=\left(4 - \frac{5 \sqrt{3}}{2},\ - \frac{1}{2}\right)

    Substituting the parameter value gives a point on the curve.

  12. Evaluate the coordinates at t=\pi

    t=π: (x,y)=(−1, −3)t=\pi:\ (x,y)=\left(-1,\ -3\right)

    Substituting the parameter value gives a point on the curve.

  13. Evaluate the coordinates at t=\frac{7 \pi}{6}

    t=7π6: (x,y)=(4−532, −112)t=\frac{7 \pi}{6}:\ (x,y)=\left(4 - \frac{5 \sqrt{3}}{2},\ - \frac{11}{2}\right)

    Substituting the parameter value gives a point on the curve.

  14. Evaluate the coordinates at t=\frac{5 \pi}{4}

    t=5π4: (x,y)=(4−522, −522−3)t=\frac{5 \pi}{4}:\ (x,y)=\left(4 - \frac{5 \sqrt{2}}{2},\ - \frac{5 \sqrt{2}}{2} - 3\right)

    Substituting the parameter value gives a point on the curve.

  15. Select the correct Cartesian equation

    (x−4)2+(y+3)2=25\left(x - 4\right)^{2} + \left(y + 3\right)^{2} = 25

    Eliminating the parameter gives this relation between x and y.

Answer
(x−4)2+(y+3)2=25\left(x - 4\right)^{2} + \left(y + 3\right)^{2} = 25
Question 2
8 markschallenging
A curve has parametric equations x=5sec⁡(t)x=5 \sec{\left(t \right)} and y=6tan⁡(t)y=6 \tan{\left(t \right)}. Which of the following is its Cartesian equation?
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Worked solution

  1. Write \sec t and \tan t in terms of x and y

    sec⁡t=x5,tan⁡t=y6\sec t=\frac{x}{5},\quad \tan t=\frac{y}{6}

    Divide each parametric equation by its constant.

  2. Apply the identity \sec^{2}t-\tan^{2}t=1

    (x5)2−(y6)2=1\left(\frac{x}{5}\right)^{2}-\left(\frac{y}{6}\right)^{2}=1

    Subtracting the squares eliminates the parameter.

  3. Recall the method

    sec⁡2t−tan⁡2t=1\sec^{2}t-\tan^{2}t=1

    This identity removes the parameter for x=a\sec t, y=b\tan t.

  4. Evaluate the coordinates at t=0

    t=0: (x,y)=(5, 0)t=0:\ (x,y)=\left(5,\ 0\right)

    Substituting the parameter value gives a point on the curve.

  5. Evaluate the coordinates at t=\frac{\pi}{6}

    t=π6: (x,y)=(1033, 23)t=\frac{\pi}{6}:\ (x,y)=\left(\frac{10 \sqrt{3}}{3},\ 2 \sqrt{3}\right)

    Substituting the parameter value gives a point on the curve.

  6. Evaluate the coordinates at t=\frac{\pi}{4}

    t=π4: (x,y)=(52, 6)t=\frac{\pi}{4}:\ (x,y)=\left(5 \sqrt{2},\ 6\right)

    Substituting the parameter value gives a point on the curve.

  7. Evaluate the coordinates at t=\frac{\pi}{3}

    t=π3: (x,y)=(10, 63)t=\frac{\pi}{3}:\ (x,y)=\left(10,\ 6 \sqrt{3}\right)

    Substituting the parameter value gives a point on the curve.

  8. Evaluate the coordinates at t=- \frac{\pi}{6}

    t=−π6: (x,y)=(1033, −23)t=- \frac{\pi}{6}:\ (x,y)=\left(\frac{10 \sqrt{3}}{3},\ - 2 \sqrt{3}\right)

    Substituting the parameter value gives a point on the curve.

  9. Evaluate the coordinates at t=- \frac{\pi}{4}

    t=−π4: (x,y)=(52, −6)t=- \frac{\pi}{4}:\ (x,y)=\left(5 \sqrt{2},\ -6\right)

    Substituting the parameter value gives a point on the curve.

  10. Evaluate the coordinates at t=- \frac{\pi}{3}

    t=−π3: (x,y)=(10, −63)t=- \frac{\pi}{3}:\ (x,y)=\left(10,\ - 6 \sqrt{3}\right)

    Substituting the parameter value gives a point on the curve.

  11. Evaluate the coordinates at t=\frac{2 \pi}{3}

    t=2π3: (x,y)=(−10, −63)t=\frac{2 \pi}{3}:\ (x,y)=\left(-10,\ - 6 \sqrt{3}\right)

    Substituting the parameter value gives a point on the curve.

  12. Evaluate the coordinates at t=\frac{3 \pi}{4}

    t=3π4: (x,y)=(−52, −6)t=\frac{3 \pi}{4}:\ (x,y)=\left(- 5 \sqrt{2},\ -6\right)

    Substituting the parameter value gives a point on the curve.

  13. Evaluate the coordinates at t=\frac{5 \pi}{6}

    t=5π6: (x,y)=(−1033, −23)t=\frac{5 \pi}{6}:\ (x,y)=\left(- \frac{10 \sqrt{3}}{3},\ - 2 \sqrt{3}\right)

    Substituting the parameter value gives a point on the curve.

  14. Evaluate the coordinates at t=\pi

    t=π: (x,y)=(−5, 0)t=\pi:\ (x,y)=\left(-5,\ 0\right)

    Substituting the parameter value gives a point on the curve.

  15. Select the correct Cartesian equation

    x225−y236=1\frac{x^{2}}{25} - \frac{y^{2}}{36} = 1

    This is the equation obtained by eliminating the parameter.

Answer
x225−y236=1\frac{x^{2}}{25} - \frac{y^{2}}{36} = 1
Question 3
8 markschallenging
A curve has parametric equations x=t2x=t^{2} and y=2t3y=2 t^{3}. Which of the following is its Cartesian equation?
Show worked solution

Worked solution

  1. Express t^{2} using the equation for x

    t2=xt^{2}=x

    The equation for x gives t^{2} directly.

  2. Square y and replace t^{2} by x

    y2=4 t6=4x3y^{2}=4\,t^{6}=4 x^{3}

    Since y^{2}=4 t^{6}=4 (t^{2})^{3}, substituting gives the Cartesian form.

  3. Recall the method

    Eliminate t using t2=x.\text{Eliminate } t \text{ using } t^{2}=x.

    Writing y^{2} in terms of t^{6}=(t^{2})^{3} removes the parameter.

  4. Evaluate the coordinates at t=-4

    t=−4: (x,y)=(16, −128)t=-4:\ (x,y)=\left(16,\ -128\right)

    Substituting the parameter value gives a point on the curve.

  5. Evaluate the coordinates at t=-3

    t=−3: (x,y)=(9, −54)t=-3:\ (x,y)=\left(9,\ -54\right)

    Substituting the parameter value gives a point on the curve.

  6. Evaluate the coordinates at t=-2

    t=−2: (x,y)=(4, −16)t=-2:\ (x,y)=\left(4,\ -16\right)

    Substituting the parameter value gives a point on the curve.

  7. Evaluate the coordinates at t=-1

    t=−1: (x,y)=(1, −2)t=-1:\ (x,y)=\left(1,\ -2\right)

    Substituting the parameter value gives a point on the curve.

  8. Evaluate the coordinates at t=0

    t=0: (x,y)=(0, 0)t=0:\ (x,y)=\left(0,\ 0\right)

    Substituting the parameter value gives a point on the curve.

  9. Evaluate the coordinates at t=1

    t=1: (x,y)=(1, 2)t=1:\ (x,y)=\left(1,\ 2\right)

    Substituting the parameter value gives a point on the curve.

  10. Evaluate the coordinates at t=2

    t=2: (x,y)=(4, 16)t=2:\ (x,y)=\left(4,\ 16\right)

    Substituting the parameter value gives a point on the curve.

  11. Evaluate the coordinates at t=3

    t=3: (x,y)=(9, 54)t=3:\ (x,y)=\left(9,\ 54\right)

    Substituting the parameter value gives a point on the curve.

  12. Evaluate the coordinates at t=4

    t=4: (x,y)=(16, 128)t=4:\ (x,y)=\left(16,\ 128\right)

    Substituting the parameter value gives a point on the curve.

  13. Evaluate the coordinates at t=5

    t=5: (x,y)=(25, 250)t=5:\ (x,y)=\left(25,\ 250\right)

    Substituting the parameter value gives a point on the curve.

  14. Evaluate the coordinates at t=6

    t=6: (x,y)=(36, 432)t=6:\ (x,y)=\left(36,\ 432\right)

    Substituting the parameter value gives a point on the curve.

  15. Select the correct Cartesian equation

    y2=4x3y^{2} = 4 x^{3}

    This is the equation obtained by eliminating the parameter.

Answer
y2=4x3y^{2} = 4 x^{3}
Question 4
8 markschallenging
The curve with parametric equations x=3tx=3 t and y=t2+1y=t^{2} + 1 meets the line y=4x−3y = 4 x - 3. Find the coordinates of the point(s) of intersection.
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Worked solution

  1. Substitute the parametric expressions into the line's equation

    t2+1=12t−3t^{2} + 1=12 t - 3

    The point of intersection lies on both the curve and the line.

  2. Rearrange into a single equation in t

    t2−12t+4=0t^{2} - 12 t + 4=0

    Collecting terms gives an equation to solve for the parameter.

  3. Solve for t

    t=6−42, 42+6t=6 - 4 \sqrt{2},\ 4 \sqrt{2} + 6

    The roots are the parameter values at the intersections.

  4. Find the coordinates for each value of t

    (18−122, 69−482), (122+18, 482+69)\left(18 - 12 \sqrt{2},\ 69 - 48 \sqrt{2}\right),\ \left(12 \sqrt{2} + 18,\ 48 \sqrt{2} + 69\right)

    Substituting each t gives an intersection point.

  5. Recall the method

    y=4x−3y=4 x - 3

    Intersection points satisfy the line equation as well.

  6. Evaluate the coordinates at t=-4

    t=−4: (x,y)=(−12, 17)t=-4:\ (x,y)=\left(-12,\ 17\right)

    Substituting the parameter value gives a point on the curve.

  7. Evaluate the coordinates at t=-3

    t=−3: (x,y)=(−9, 10)t=-3:\ (x,y)=\left(-9,\ 10\right)

    Substituting the parameter value gives a point on the curve.

  8. Evaluate the coordinates at t=-2

    t=−2: (x,y)=(−6, 5)t=-2:\ (x,y)=\left(-6,\ 5\right)

    Substituting the parameter value gives a point on the curve.

  9. Evaluate the coordinates at t=-1

    t=−1: (x,y)=(−3, 2)t=-1:\ (x,y)=\left(-3,\ 2\right)

    Substituting the parameter value gives a point on the curve.

  10. Evaluate the coordinates at t=0

    t=0: (x,y)=(0, 1)t=0:\ (x,y)=\left(0,\ 1\right)

    Substituting the parameter value gives a point on the curve.

  11. Evaluate the coordinates at t=1

    t=1: (x,y)=(3, 2)t=1:\ (x,y)=\left(3,\ 2\right)

    Substituting the parameter value gives a point on the curve.

  12. Evaluate the coordinates at t=2

    t=2: (x,y)=(6, 5)t=2:\ (x,y)=\left(6,\ 5\right)

    Substituting the parameter value gives a point on the curve.

  13. Evaluate the coordinates at t=3

    t=3: (x,y)=(9, 10)t=3:\ (x,y)=\left(9,\ 10\right)

    Substituting the parameter value gives a point on the curve.

  14. Evaluate the coordinates at t=4

    t=4: (x,y)=(12, 17)t=4:\ (x,y)=\left(12,\ 17\right)

    Substituting the parameter value gives a point on the curve.

  15. State the point(s) of intersection

    (18−122, 69−482), (122+18, 482+69)\left(18 - 12 \sqrt{2},\ 69 - 48 \sqrt{2}\right),\ \left(12 \sqrt{2} + 18,\ 48 \sqrt{2} + 69\right)

    These points lie on both the curve and the line.

Answer
(18−122, 69−482), (122+18, 482+69)\left(18 - 12 \sqrt{2},\ 69 - 48 \sqrt{2}\right),\ \left(12 \sqrt{2} + 18,\ 48 \sqrt{2} + 69\right)
Question 5
8 markschallenging
The curve with parametric equations x=t−1x=t - 1 and y=t2y=t^{2} meets the line y=4x+3y = 4 x + 3. Find the coordinates of the point(s) of intersection.
Show worked solution

Worked solution

  1. Substitute the parametric expressions into the line's equation

    t2=4t−1t^{2}=4 t - 1

    The point of intersection lies on both the curve and the line.

  2. Rearrange into a single equation in t

    t2−4t+1=0t^{2} - 4 t + 1=0

    Collecting terms gives an equation to solve for the parameter.

  3. Solve for t

    t=2−3, 3+2t=2 - \sqrt{3},\ \sqrt{3} + 2

    The roots are the parameter values at the intersections.

  4. Find the coordinates for each value of t

    (1−3, 7−43), (1+3, 43+7)\left(1 - \sqrt{3},\ 7 - 4 \sqrt{3}\right),\ \left(1 + \sqrt{3},\ 4 \sqrt{3} + 7\right)

    Substituting each t gives an intersection point.

  5. Recall the method

    y=4x+3y=4 x + 3

    Intersection points satisfy the line equation as well.

  6. Evaluate the coordinates at t=-4

    t=−4: (x,y)=(−5, 16)t=-4:\ (x,y)=\left(-5,\ 16\right)

    Substituting the parameter value gives a point on the curve.

  7. Evaluate the coordinates at t=-3

    t=−3: (x,y)=(−4, 9)t=-3:\ (x,y)=\left(-4,\ 9\right)

    Substituting the parameter value gives a point on the curve.

  8. Evaluate the coordinates at t=-2

    t=−2: (x,y)=(−3, 4)t=-2:\ (x,y)=\left(-3,\ 4\right)

    Substituting the parameter value gives a point on the curve.

  9. Evaluate the coordinates at t=-1

    t=−1: (x,y)=(−2, 1)t=-1:\ (x,y)=\left(-2,\ 1\right)

    Substituting the parameter value gives a point on the curve.

  10. Evaluate the coordinates at t=0

    t=0: (x,y)=(−1, 0)t=0:\ (x,y)=\left(-1,\ 0\right)

    Substituting the parameter value gives a point on the curve.

  11. Evaluate the coordinates at t=1

    t=1: (x,y)=(0, 1)t=1:\ (x,y)=\left(0,\ 1\right)

    Substituting the parameter value gives a point on the curve.

  12. Evaluate the coordinates at t=2

    t=2: (x,y)=(1, 4)t=2:\ (x,y)=\left(1,\ 4\right)

    Substituting the parameter value gives a point on the curve.

  13. Evaluate the coordinates at t=3

    t=3: (x,y)=(2, 9)t=3:\ (x,y)=\left(2,\ 9\right)

    Substituting the parameter value gives a point on the curve.

  14. Evaluate the coordinates at t=4

    t=4: (x,y)=(3, 16)t=4:\ (x,y)=\left(3,\ 16\right)

    Substituting the parameter value gives a point on the curve.

  15. State the point(s) of intersection

    (1−3, 7−43), (1+3, 43+7)\left(1 - \sqrt{3},\ 7 - 4 \sqrt{3}\right),\ \left(1 + \sqrt{3},\ 4 \sqrt{3} + 7\right)

    These points lie on both the curve and the line.

Answer
(1−3, 7−43), (1+3, 43+7)\left(1 - \sqrt{3},\ 7 - 4 \sqrt{3}\right),\ \left(1 + \sqrt{3},\ 4 \sqrt{3} + 7\right)

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