All subjects · Further Mathematics · WFM01 Further Pure Mathematics 1

Tangents and normals to a parabola

Edexcel IAL Further Mathematics · WFM01 Further Pure Mathematics 1

To find a tangent or normal to y2 = 4ax you need the gradient. In F1 you get it by differentiating the Cartesian form (parametric differentiation is not required).

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The gradient

For y > 0, y = 2sqrt(a) x1/2, so dy/dx = sqrt(a) x-1/2 = sqrt(a)/sqrt(x).

At the point (at2, 2at) with t > 0, sqrt(x) = sqrt(a) t, so dy/dx = 1/t. The same result holds for t < 0 using y = -2sqrt(ax). (If you know implicit differentiation, 2y dy/dx = 4a gives dy/dx = 2a/y = 1/t directly.)

Tangent at (at2, 2at)

y - 2at = (1/t)(x - at2). Multiply by t: ty - 2at2 = x - at2, so ty = x + at2.

It meets the x-axis at (-at2, 0) and the y-axis at (0, at).

Normal at (at2, 2at)

The normal is perpendicular to the tangent, so its gradient is -t:

y - 2at = -t(x - at2), so y + tx = 2at + at3.

In exam questions, simplify the gradient before substituting, and watch brackets: (2at)2 = 4a2t2, not 2at2.

Worked example

Find the equation of the normal to y2 = 12x at the point (12, 12).

  1. a = 3 and 12 = 2(3)t gives t = 2. Check x = 3(22) = 12.
  2. Gradient of tangent = 1/t = 1/2, so gradient of normal = -2.
  3. y - 12 = -2(x - 12)

Answer: y = -2x + 36

Key idea: On y2 = 4ax at (at2, 2at): tangent gradient 1/t, tangent ty = x + at2, normal y + tx = 2at + at3.

Check yourself

  1. Find the gradient of the tangent to y2 = 12x at the point (12, 12). Give your answer as a fraction in its simplest form.
    Show answer

    1/2. a = 3 and t = 2, so dy/dx = 1/t = 1/2. (Or y = 2sqrt(3) x1/2, dy/dx = sqrt(3)/sqrt(x) = sqrt(3)/sqrt(12) = 1/2.)

  2. The tangent to y2 = 12x at (3, 6) has equation y = x + k. Find k.
    Show answer

    3. a = 3 and t = 1. ty = x + at2 gives y = x + 3.

  3. The normal to y2 = 12x at (12, 12) meets the y-axis at (0, k). Find k.
    Show answer

    36. The normal is y + 2x = 2(3)(2) + 3(8) = 36, so k = 36.

  4. The tangent to y2 = 8x at the point with parameter t = 3 meets the x-axis at (k, 0). Find k.
    Show answer

    -18. The tangent is 3y = x + 18. Setting y = 0 gives x = -18 = -at2.

  5. Which is the equation of the tangent to y2 = 4ax at (at2, 2at)?
    • y + tx = 2at + at3
    • ty = x + at2
    • ty = x - at2
    • y = tx + at2
    Show answer

    ty = x + at2. Gradient 1/t: y - 2at = (1/t)(x - at2), then multiply by t.

  6. Find the gradient of the normal to y2 = 4ax at the point with parameter t = -2.
    Show answer

    2. The tangent gradient is 1/t = -1/2, so the normal gradient is -t = 2.

  7. The tangent to y2 = 20x at the point with parameter t = 2 meets the y-axis at (0, k). Find k.
    Show answer

    10. a = 5. The tangent is 2y = x + 20, so at x = 0, y = 10 = at.

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Last updated 8 November 2026. Spotted a mistake? Tell us.