01 · A tangent
Find the tangent to y = x² − 2x + 3 at x = 2.
Hint
The point is (2,3); the gradient is 2.
Worked solution
y − 3 = 2(x − 2)
y = 2x − 1.
Understand · explore · practise
Tangents and normals
Find tangent and normal equations using derivatives. Handle horizontal tangents, intersections, external-point tangents and second meetings with a curve.
Before you startDifferentiation rules, equations of lines and simultaneous equations
01 / Find a tangent
The curve supplies the point; its derivative supplies the slope.
At P(a, f(a)):
y − f(a) = f′(a)(x − a)
For f(x) = x² + x − 2 at a = 1, P = (1,0) and f′(1) = 3. The tangent is y = 3(x − 1), or y = 3x − 3.
Move the point in the model. The blue curve gives its height, while the tangent and normal pass through that same point. Equal horizontal and vertical scales make perpendicularity meaningful in the drawing.
Move the point of contactMove at your pace
At P = (1,0), the tangent slope is 3. The normal slope is -0.3333; the product of the slopes is −1. Blue is the curve, gold the tangent and green the normal. The coordinate scales are equal.
Watch the tangent and normal change together
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Find a normal
The normal is perpendicular to the tangent at the point.
mnormal = −1/f′(a), if f′(a) is finite and non-zero
At P(1,0) on the same curve, the tangent slope is 3, so the normal slope is −1/3:
y = −(1/3)(x − 1)
x + 3y − 1 = 0
Use the original point on the curve, not an intercept of the tangent. As a quick check, the two finite non-zero slopes multiply to −1.
03 / Horizontal and vertical cases
A zero tangent slope gives a vertical normal.
For f(x) = x² + x − 2, f′(x) = 2x + 1 is zero at x = −1/2. The point is (−1/2,−9/4).
Horizontal tangent: y = −9/4
Vertical normal: x = −1/2
Do not try to calculate −1/0. A vertical line is written x = constant and has no finite gradient. If a curve instead has a vertical tangent, its normal can be horizontal; the usual finite-derivative tangent formula needs separate treatment.
04 / A reliable order
Check the point before calculating either line.
Four checksWorked example
1. Find or verify P(a, f(a)).
If coordinates are supplied, substitute them into the curve.
2. Differentiate, then evaluate f′(a).
The derivative is not usually the y-coordinate.
3. Choose tangent or normal slope.
For a normal, handle a zero tangent slope separately.
4. Use y − y₀ = m(x − x₀).
Substitute P into the final line to check it passes through the point.
A root function
For y = 6 − 2√x at x = 4, the point is (4,2) and dy/dx = −1/√x = −1/2. The normal slope is 2, so its equation is y − 2 = 2(x − 4), or y = 2x − 6.
05 / Intersect two lines
Different points on the curve give different normals.
On y = x² + x − 2, the normals at A(0,−2) and B(2,4) have slopes −1 and −1/5:
At A: y = −x − 2
At B: y = −x/5 + 22/5
Equating them gives −5x − 10 = −x + 22, so they meet at N(−8,6).
The triangle enclosed by the points
AB = (2,6) and AN = (−8,8). The triangle area is half the absolute determinant:
Area ABN = ½|2 × 8 − 6 × (−8)| = 32
Alternatively use a coordinate-area method. A sketch helps distinguish the triangle’s sides from the infinite normal lines.
06 / A tangent through an external point
The unknown is the point of contact.
Find tangents to y = x² + 2 that pass through (0,−2). Write the contact point as (a,a² + 2), with tangent slope 2a.
y − (a² + 2) = 2a(x − a)
y = 2ax − a² + 2
Putting (0,−2) into the tangent gives a² = 4, so a = ±2. The two tangents are y = 4x − 2 and y = −4x − 2. If a positive gradient is specified, choose the first.
Check using the discriminant
A line y = mx − 2 meets the parabola where x² − mx + 4 = 0. Tangency gives a repeated root: m² − 16 = 0, so m = ±4, agreeing with the derivative method.
07 / Where does the line meet the curve again?
Keep the known contact root while solving the intersection.
The normal to y = x² at P(1,1) is y − 1 = −(x − 1)/2. Substituting y = x² gives:
2x² + x − 3 = 0
(x − 1)(2x + 3) = 0
The known root x = 1 gives P. The other root x = −3/2 gives Q(−3/2,9/4).
A tangent can meet the curve elsewhere
For g(x) = x³ + x² − 2x − 1, the tangent at P(0,−1) is y = −2x − 1. Their intersection equation is x³ + x² = x²(x + 1) = 0. Besides the contact root x = 0, there is Q(−1,1). Thus PQ has length √5.
The repeated contact root fits this polynomial example; do not assume every curve-line problem is a quadratic discriminant calculation.
08 / Your turn
Give a line equation, not just a gradient.
Find exact equations and coordinates.
02 · A normal
Find the normal at the same point.
Hint
Use slope −1/2.
Worked solution
y − 3 = −(x − 2)/2
x + 2y − 8 = 0.
03 · A stationary point
Find the tangent and normal to y = x² − 2x + 3 at x = 1.
Hint
The derivative is zero and the point is (1,2).
Worked solution
Tangent: y = 2
Normal: x = 1.
04 · Verify the point
Why is “the tangent to y = x² + 1 at (2,4)” not a valid request as written?
Hint
Evaluate the curve at x = 2.
Worked solution
The curve has y = 5 there, so (2,4) is not on it. If the intended point is (2,5), the tangent is y − 5 = 4(x − 2), or y = 4x − 3.
05 · A reciprocal curve
Find the tangent and normal to y = 4/x at x = 2.
Hint
The point is (2,2), with derivative −4/x².
Worked solution
Tangent slope −1: y = −x + 4
Normal slope 1: y = x.
06 · Tangent intercepts
The tangent y = −x + 4 cuts the axes at R and S. Find RS and the triangle area with the origin.
Hint
The intercepts are (4,0) and (0,4).
Worked solution
RS = √(16 + 16) = 4√2
Triangle area = ½ × 4 × 4 = 8.
07 · Different points
Find where the tangent to y = x² + x − 2 at x = 1 meets the normal at x = 0.
Hint
Solve y = 3x − 3 and y = −x − 2.
Worked solution
4x = 1 ⇒ x = 1/4
y = −9/4.
08 · Two external tangents
Find both tangents to y = x² that pass through (0,−9).
Hint
A tangent at x = a has equation y = 2ax − a².
Worked solution
a² = 9 ⇒ a = ±3
y = 6x − 9 or y = −6x − 9.
09 · A normal’s second meeting
Find the second intersection of y = x² with its normal at (2,4).
Hint
The normal has slope −1/4. Substitute y = x² into its equation.
Worked solution
y − 4 = −(x − 2)/4
4x² + x − 18 = (x − 2)(4x + 9) = 0
Q = (−9/4, 81/16).
The root x = 2 is the known starting point.
09 / Recap
Point, slope, line, check.
- Evaluate f for the point and f′ for its tangent slope.
- Use the negative reciprocal for a normal only when valid.
- A horizontal tangent has a vertical normal.
- Different contact points can give intersecting normal lines.
- For an external tangent, solve for the unknown point of contact.
- Check all curve-line intersections, including the known contact point.
Section 1 of 9 · Find a tangent