Tangents and normals: 7 exam questions solved step by step

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What this lesson covers

  • The derivative gives the tangent's gradient; the normal's gradient is its negative reciprocal.
  • An equation is finished only when it is in y = mx + c or general form.
Key idea

7 questions, one method

  • 1. The point: both coordinates.
  • 2. The gradient: dy/dx at that x.
  • 3. Normal: the negative reciprocal.
  • 4. Finish: y = mx + c or general form.
Your turn

Question 1 · Achieved

  • Find the equation of the tangent to
  • y = x³ − 2x² + 3 at x = 2.
Worked example

Q1: tangent at x = 2

  • Point, then gradient.
  • Finish as y = mx + c.
  1. x = 2: y = 8 − 8 + 3 = 3the point
  2. dy/dx = 3x² − 4xdifferentiate
  3. x = 2: m = 12 − 8 = 4the gradient
  4. y − 3 = 4(x − 2)point-gradient form
  5. y − 3 = 4x − 8expand
  6. y = 4x − 5add 3
  7. x = 2: 4(2) − 5 = 3through (2, 3)
Your turn

Question 2 · Achieved

  • Find the equation of the normal to
  • y = x + 2sin x at x = π/2.
  • Careful: normal, not tangent.
Worked example

Q2: normal at x = π/2

  • Radians throughout.
  • Normal m = −1 ÷ m.
  1. x = π/2: y = π/2 + 2sin(π/2)
  2. = π/2 + 2sin(π/2) = 1
  3. dy/dx = 1 + 2cos xd/dx sin x = cos x
  4. x = π/2: m = 1 + 2(0) = 1cos(π/2) = 0
  5. y = x + 2that is the tangent
  6. normal m = −1−1 ÷ 1
  7. y − (π/2 + 2) = −(x − π/2)
  8. y = −x + π/2 + π/2 + 2expand
  9. y = −x + π + 2π/2 + π/2 = π
Your turn

Question 3 · Merit

  • y = x³ − 6x² + 5
  • Find the points where the tangent
  • is parallel to y = −9x + 2.
Worked example

Q3: parallel to y = −9x + 2

  • Parallel: same gradient.
  • Find both y values.
  1. parallel: m = −9same gradient
  2. dy/dx = 3x² − 12xdifferentiate
  3. 3x² − 12x = −9gradient −9
  4. 3x² − 12x + 9 = 0add 9
  5. x² − 4x + 3 = 0÷ 3
  6. (x − 1)(x − 3) = 0factorise
  7. x = 1: y = 1 − 6 + 5 = 0
  8. x = 3: y = 27 − 54 + 5 = −22
  9. (1, 0) and (3, −22)
Your turn

Question 4 · Merit

  • The tangent to y = ln(2x − 3) at x = 2
  • forms a triangle with the two axes.
  • Find the area of the triangle.
Worked example

Q4: the tangent first

  • ln: use the chain rule.
  • ln 1 = 0
  1. x = 2: y = ln(4 − 3) = ln 1 = 0
  2. dy/dx = 2/(2x − 3)chain rule
  3. x = 2: m = 2/(4 − 3) = 2the gradient
  4. y − 0 = 2(x − 2)point-gradient form
  5. y = 2x − 4expand
Worked example

Q4: where it meets the axes

  • x = 0, then y = 0.
  • A length is positive.
  1. y = 2x − 4
  2. x = 0: y = −4meets the y-axis
  3. y = 0: 2x = 4, x = 2meets the x-axis
  4. height = −4a length is positive
  5. base 2, height 4sides along the axes
  6. area = (1/2) × 2 × 4half base × height
  7. = 4 units²
Your turn

Question 5 · Merit

  • Find the normal to y = (2x + 1)/(x − 1)
  • at x = 2, in general form.
Worked example

Q5: normal to a quotient

  • Quotient rule.
  • Watch the sign of −1 ÷ m.
  1. x = 2: y = 5/1 = 5the point
  2. dy/dx = (vu′ − uv′)/v²quotient rule
  3. = (2(x − 1) − (2x + 1))/(x − 1)²
  4. = −3/(x − 1)²2x − 2 − 2x − 1
  5. x = 2: m = −3/1 = −3tangent gradient
  6. normal m = 1/3−1 ÷ −3
  7. y − 5 = (1/3)(x − 2)point-gradient form
  8. 3y − 15 = x − 2× 3
  9. x − 3y + 13 = 0general form
Your turn

Question 6 · Excellence

  • Find the equations of the two tangents
  • to y = x² + 1 that pass through
  • the point (1, −2).
Worked example

Q6: a tangent at an unknown point

  • Is (1, −2) on the curve?
  • Call the point (a, a² + 1).
  1. 1² + 1 = 2 ≠ −2(1, −2) is off the curve
  2. point of contact (a, a² + 1)
  3. dy/dx = 2x, so m = 2agradient at x = a
  4. y − (a² + 1) = 2a(x − a)tangent at a
  5. y = 2ax − 2a² + a² + 1expand
  6. y = 2ax − a² + 1
  7. −2 = 2a − a² + 1through (1, −2)
  8. a² − 2a − 3 = 0rearrange
Worked example

Q6: two values of a

  • Each a gives a tangent.
  • Check both pass through.
  1. a² − 2a − 3 = 0
  2. (a − 3)(a + 1) = 0factorise
  3. a = 3 or a = −1
  4. a = 3: y = 6x − 9 + 1into y = 2ax − a² + 1
  5. y = 6x − 8
  6. a = −1: y = −2x − 1 + 1
  7. y = −2x
  8. 6(1) − 8 = −2through (1, −2)
  9. −2(1) = −2through (1, −2)
Your turn

Question 7 · Excellence

  • The tangent to y = e³ˣ at x = a
  • passes through the origin.
  • Find a and the equation of the tangent.
Worked example

Q7: the tangent at x = a

  • Write it in terms of a.
  • Then use (0, 0).
  1. point (a, e³ᵃ)point of contact
  2. dy/dx = 3e³ˣchain rule
  3. m = 3e³ᵃat x = a
  4. y − e³ᵃ = 3e³ᵃ(x − a)tangent at a
  5. 0 − e³ᵃ = 3e³ᵃ(0 − a)through (0, 0)
  6. −e³ᵃ = −3ae³ᵃ
  7. 3ae³ᵃ − e³ᵃ = 0rearrange
  8. e³ᵃ(3a − 1) = 0factorise
Worked example

Q7: solve for a, then finish

  • e to any power is positive.
  • Finish the equation.
  1. e³ᵃ(3a − 1) = 0
  2. e³ᵃ = 0e³ᵃ > 0 for every a
  3. 3a − 1 = 0, so a = 1/3
  4. a = 1/3: y = e¹ = epoint (1/3, e)
  5. m = 3e¹ = 3e
  6. y − e = 3e(x − 1/3)point-gradient form
  7. y = 3ex − e + eexpand
  8. y = 3ex
  9. (0, 0): 3e(0) = 0
Summary

The method that solved all seven

  • Point, gradient, line, then finish it.
StepHow
Pointboth coordinates first
Gradientdy/dx at that x
Normal−1 ÷ tangent gradient
Finishy = mx + c or general form
Parallelset dy/dx = m and solve
Off curveuse (a, f(a)), solve for a

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