All subjects · Further Mathematics · WFM03 Further Pure Mathematics 3

Tangents and normals to conics

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

To find tangents and normals to an ellipse or hyperbola you need the gradient, from implicit or parametric differentiation. Then use y - y1 = m(x - x1). Learning the general results saves time.

Practise this in the app: instant marking and a fix for every mistake. Free, no email needed to start.

Practise this lesson

Finding the gradient

Implicitly: from x2/a2 + y2/b2 = 1, 2x/a2 + (2y/b2)(dy/dx) = 0, so dy/dx = -b2x/(a2y).

For the hyperbola, dy/dx = b2x/(a2y).

Parametrically: dy/dx = (dy/dt)/(dx/dt). For (a cos t, b sin t): dy/dx = (b cos t)/(-a sin t) = -(b cos t)/(a sin t).

General results for the ellipse

Tangent at (x1, y1): x x1/a2 + y y1/b2 = 1.

Tangent at (a cos t, b sin t): bx cos t + ay sin t = ab.

Normal at (a cos t, b sin t): ax sin t - by cos t = (a2 - b2) sin t cos t.

The normal has gradient -1/m, where m is the tangent gradient.

General results for the hyperbola

Tangent at (x1, y1): x x1/a2 - y y1/b2 = 1.

Tangent at (a sec t, b tan t): bx sec t - ay tan t = ab.

Tangent at (a cosh t, b sinh t): bx cosh t - ay sinh t = ab.

In the exam you may be asked to derive these, so practise the method, not just the formula.

Worked example

Find the equation of the tangent to x2/16 + y2/9 = 1 at the point with parameter t (x = 4 cos t, y = 3 sin t).

  1. dx/dt = -4 sin t, dy/dt = 3 cos t, so dy/dx = -3cos t/(4 sin t)
  2. y - 3 sin t = -(3cos t/(4 sin t))(x - 4 cos t)
  3. Multiply by 4 sin t: 4y sin t - 12sin2 t = -3x cos t + 12cos2 t
  4. Use sin2 t + cos2 t = 1

Answer: 3x cos t + 4y sin t = 12

Key idea: Ellipse gradient -b2x/(a2y); hyperbola b2x/(a2y). Tangent at (x1, y1): x x1/a2 ± y y1/b2 = 1. Normal gradient = -1/(tangent gradient).

Check yourself

  1. Find the gradient of the tangent to x2/16 + y2/9 = 1 at (2, 3sqrt(3)/2), to 3 decimal places.
    Show answer

    -0.433. dy/dx = -9x/(16y) = -18/(24sqrt(3)) = -sqrt(3)/4 = -0.433.

  2. What is the equation of the tangent to x2/a2 + y2/b2 = 1 at (a cos t, b sin t)?
    • ax cos t + by sin t = ab
    • bx sin t + ay cos t = ab
    • bx cos t + ay sin t = ab
    • bx cos t - ay sin t = ab
    Show answer

    bx cos t + ay sin t = ab. Equivalent to x cos t/a + y sin t/b = 1; multiply through by ab.

  3. Find the y-intercept of the tangent to x2/4 + y2 = 1 at (sqrt(3), 1/2).
    Show answer

    2. Tangent: sqrt(3)x/4 + y/2 = 1. At x = 0, y/2 = 1, so y = 2.

  4. The point (5, 8/3) lies on x2/9 - y2/4 = 1. Find the gradient of the normal there.
    Show answer

    -1.2. dy/dx = b2x/(a2y) = 4(5)/(9 × 8/3) = 20/24 = 5/6, so the normal has gradient -6/5 = -1.2.

  5. Which is the tangent to x2/a2 - y2/b2 = 1 at the point (x1, y1)?
    • x x1/a2 + y y1/b2 = 1
    • x x1/a2 - y y1/b2 = 1
    • x x1/a2 - y y1/b2 = 0
    • x/a2 - y/b2 = x1 - y1
    Show answer

    x x1/a2 - y y1/b2 = 1. Implicit differentiation gives dy/dx = b2x1/(a2y1); rearranging the tangent and using the curve equation gives this form.

  6. The tangent to x2/25 + y2/9 = 1 at the point with parameter t = π/3 crosses the x-axis at A. Find the x-coordinate of A.
    Show answer

    10. Tangent: 3x cos t + 5y sin t = 15. At y = 0, x = 5/cos t = 5 ÷ 1/2 = 10.

  7. The tangent to a conic at P has gradient m (m ≠ 0). What is the gradient of the normal at P?
    • 1/m
    • -1/m
    • -m
    • m2
    Show answer

    -1/m. The normal is perpendicular to the tangent, so the gradients multiply to -1.

Practise with instant feedback

Sources

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