Parametric curves Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Parametric curves questions. See exactly how to solve problems on parametric, cartesian, linear, parabola.

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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.

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=t−2x=t - 2 and y=2t+3y=2 t + 3, where tt is a parameter. Find a Cartesian equation of the curve.

Worked solution

  1. Make t the subject of the equation for x

    t=x+21t=\frac{x + 2}{1}

    Rearranging the linear equation for x isolates the parameter.

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

    y=2(x+21)+3y=2\left(\frac{x + 2}{1}\right)+3

    Replacing t leaves an equation in x and y only.

  3. State the Cartesian equation of the curve

    y=2x+7y = 2 x + 7

    This equation contains no parameter, as required.

Answer
y=2x+7y = 2 x + 7
Question 3
2 markseasy
A curve has parametric equations x=3t+2x=3 t + 2 and y=t+4y=t + 4, where tt is a parameter. Find a Cartesian equation of the curve.

Worked solution

  1. Make t the subject of the equation for x

    t=x−23t=\frac{x - 2}{3}

    Rearranging the linear equation for x isolates the parameter.

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

    y=1(x−23)+4y=1\left(\frac{x - 2}{3}\right)+4

    Replacing t leaves an equation in x and y only.

  3. State the Cartesian equation of the curve

    y=x3+103y = \frac{x}{3} + \frac{10}{3}

    This equation contains no parameter, as required.

Answer
y=x3+103y = \frac{x}{3} + \frac{10}{3}
Question 4
2 markseasy
A curve has parametric equations x=2t−3x=2 t - 3 and y=4t+1y=4 t + 1, where tt is a parameter. Find a Cartesian equation of the curve.

Worked solution

  1. Make t the subject of the equation for x

    t=x+32t=\frac{x + 3}{2}

    Rearranging the linear equation for x isolates the parameter.

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

    y=4(x+32)+1y=4\left(\frac{x + 3}{2}\right)+1

    Replacing t leaves an equation in x and y only.

  3. State the Cartesian equation of the curve

    y=2x+7y = 2 x + 7

    This equation contains no parameter, as required.

Answer
y=2x+7y = 2 x + 7
Question 5
2 markseasy
A curve has parametric equations x=t+1x=t + 1 and y=t2y=t^{2}, where tt is a parameter. Find a Cartesian equation of the curve.

Worked solution

  1. Make t the subject of the equation for x

    t=x−1t=x - 1

    The equation for x is linear in t, so t is easily isolated.

  2. Substitute t=x-1 into y=t^{2}+0

    y=(x−1)2+0y=\left(x - 1\right)^{2}+0

    Squaring the linear expression gives a quadratic in x.

  3. State the Cartesian equation of the curve

    y=x2−2x+1y = x^{2} - 2 x + 1

    This equation contains no parameter, as required.

Answer
y=x2−2x+1y = x^{2} - 2 x + 1

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