A-Level Parametric curves Practice Questions

Free A-Level Parametric curves practice questions with full step-by-step worked solutions. Covers parametric, cartesian, linear, parabola. Practise exam-style problems and check your method.

parametriccartesianlinearparabolareciprocalcoordinate
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A curve has parametric equations x=2t+1x=2 t + 1 and y=3t−1y=3 t - 1, where tt is a parameter. Find a Cartesian equation of the curve.
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Worked solution

  1. Make t the subject of the equation for x

    t=x−12t=\frac{x - 1}{2}

    Rearranging the linear equation for x isolates the parameter.

  2. Substitute this expression for t into the equation for y

    y=3(x−12)+−1y=3\left(\frac{x - 1}{2}\right)+-1

    Replacing t leaves an equation in x and y only.

  3. State the Cartesian equation of the curve

    y=3x2−52y = \frac{3 x}{2} - \frac{5}{2}

    This equation contains no parameter, as required.

Answer
y=3x2−52y = \frac{3 x}{2} - \frac{5}{2}
Question 2
2 markseasy
A curve has parametric equations x=tx=t and y=4ty=\frac{4}{t}. Which of the following is its Cartesian equation?
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Worked solution

  1. Make t the subject of the equation for x

    t=x1t=\frac{x}{1}

    Dividing by the coefficient of t isolates the parameter.

  2. Substitute this into y=\dfrac{4}{t}

    y=4x1y=\frac{4}{\frac{x}{1}}

    A reciprocal in t becomes a reciprocal in x.

  3. Select the correct Cartesian equation

    y=4xy = \frac{4}{x}

    This is the equation obtained by eliminating the parameter.

Answer
y=4xy = \frac{4}{x}
Question 3
3 marksintermediate
A curve has parametric equations x=3cos⁡(t)+2x=3 \cos{\left(t \right)} + 2 and y=3sin⁡(t)+1y=3 \sin{\left(t \right)} + 1. Which of the following is its Cartesian equation?
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Worked solution

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

    cos⁡t=x−23,sin⁡t=y−13\cos t=\frac{x - 2}{3},\quad \sin t=\frac{y - 1}{3}

    Rearrange each parametric equation for the trig ratio.

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

    (x−23)2+(y−13)2=1\left(\frac{x - 2}{3}\right)^{2}+\left(\frac{y - 1}{3}\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)=(5, 1)t=0:\ (x,y)=\left(5,\ 1\right)

    Substituting the parameter value gives a point on the curve.

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

    t=π6: (x,y)=(2+332, 52)t=\frac{\pi}{6}:\ (x,y)=\left(2 + \frac{3 \sqrt{3}}{2},\ \frac{5}{2}\right)

    Substituting the parameter value gives a point on the curve.

  6. Select the correct Cartesian equation

    (x−2)2+(y−1)2=9\left(x - 2\right)^{2} + \left(y - 1\right)^{2} = 9

    This is the equation obtained by eliminating the parameter.

Answer
(x−2)2+(y−1)2=9\left(x - 2\right)^{2} + \left(y - 1\right)^{2} = 9
Question 4
5 markshard
The curve x=4sec⁡(t)x=4 \sec{\left(t \right)}, y=3tan⁡(t)y=3 \tan{\left(t \right)} can be written in Cartesian form. Select the correct equation.
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Worked solution

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

    sec⁡t=x4,tan⁡t=y3\sec t=\frac{x}{4},\quad \tan t=\frac{y}{3}

    Divide each parametric equation by its constant.

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

    (x4)2−(y3)2=1\left(\frac{x}{4}\right)^{2}-\left(\frac{y}{3}\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)=(4, 0)t=0:\ (x,y)=\left(4,\ 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)=(833, 3)t=\frac{\pi}{6}:\ (x,y)=\left(\frac{8 \sqrt{3}}{3},\ \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)=(42, 3)t=\frac{\pi}{4}:\ (x,y)=\left(4 \sqrt{2},\ 3\right)

    Substituting the parameter value gives a point on the curve.

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

    t=π3: (x,y)=(8, 33)t=\frac{\pi}{3}:\ (x,y)=\left(8,\ 3 \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)=(833, −3)t=- \frac{\pi}{6}:\ (x,y)=\left(\frac{8 \sqrt{3}}{3},\ - \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)=(42, −3)t=- \frac{\pi}{4}:\ (x,y)=\left(4 \sqrt{2},\ -3\right)

    Substituting the parameter value gives a point on the curve.

  10. Select the correct Cartesian equation

    x216−y29=1\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1

    Eliminating the parameter gives this relation between x and y.

Answer
x216−y29=1\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1
Question 5
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

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