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This lesson focuses on finding where curves intersect — lines meeting curves, two curves meeting each other, and how the discriminant determines the nature of the intersection. These problems bring together algebra and coordinate geometry and are a staple of A-Level examinations.
To find where a straight line meets a curve, substitute the equation of the line into the equation of the curve. This produces an equation in one variable, which you solve.
Example 1: Find where the line y = 2x − 1 meets the parabola y = x².
Set equal: x² = 2x − 1
x² − 2x + 1 = 0
(x − 1)² = 0
x = 1 (repeated root)
y = 2(1) − 1 = 1
The line meets the parabola at exactly one point, (1, 1). The repeated root tells us the line is a tangent to the parabola.
Example 2: Find where y = x + 3 meets y = x² − x − 2.
x + 3 = x² − x − 2
x² − 2x − 5 = 0
x = (2 ± √(4 + 20)) / 2 = (2 ± √24) / 2 = 1 ± √6
Points: (1 + √6, 4 + √6) and (1 − √6, 4 − √6)
When a line y = mx + c is substituted into a quadratic curve, the resulting equation is a quadratic in x. The discriminant determines the number of intersection points:
| Discriminant | Meaning |
|---|---|
| Δ > 0 | Two distinct intersection points |
| Δ = 0 | One repeated point — the line is a tangent |
| Δ < 0 | No intersection points |
Example 3: Find the values of k for which the line y = kx + 1 is a tangent to the curve y = x² + 2.
Substitute: x² + 2 = kx + 1
x² − kx + 1 = 0
For tangency: Δ = 0
k² − 4(1)(1) = 0
k² = 4
k = ±2
When k = 2: x² − 2x + 1 = 0, so x = 1, y = 3. Tangent at (1, 3). When k = −2: x² + 2x + 1 = 0, so x = −1, y = 3. Tangent at (−1, 3) (by symmetry of the parabola).
Example 4: Find the range of values of c for which the line y = 3x + c does not meet the curve y = x² + 5x + 2.
x² + 5x + 2 = 3x + c
x² + 2x + (2 − c) = 0
No intersection: Δ < 0
4 − 4(2 − c) < 0
4 − 8 + 4c < 0
4c < 4
c < 1
This is a particularly common A-Level problem type.
Example 5: Find where the line y = x − 2 meets the circle x² + y² = 20.
x² + (x − 2)² = 20
x² + x² − 4x + 4 = 20
2x² − 4x − 16 = 0
x² − 2x − 8 = 0
(x − 4)(x + 2) = 0
x = 4 or x = −2
Points: (4, 2) and (−2, −4)
Example 6: Show that the line y = 2x + k is a tangent to x² + y² = 5 when k = ±5.
x² + (2x + k)² = 5
x² + 4x² + 4kx + k² = 5
5x² + 4kx + (k² − 5) = 0
Tangency: Δ = 0
16k² − 4(5)(k² − 5) = 0
16k² − 20k² + 100 = 0
−4k² + 100 = 0
k² = 25
k = ±5 ✓
Geometric interpretation: The perpendicular distance from the centre (0, 0) to the line 2x − y + k = 0 is |k|/√5. For tangency, this equals the radius √5:
|k|/√5 = √5
|k| = 5
k = ±5 ✓
To find where two curves intersect, eliminate one variable between their equations.
Example 7: Find the intersection points of y = x² and y = 4x − x².
x² = 4x − x²
2x² − 4x = 0
2x(x − 2) = 0
x = 0 or x = 2
At x = 0: y = 0. At x = 2: y = 4.
Points: (0, 0) and (2, 4).
Example 8: Find where the circles x² + y² = 25 and (x − 7)² + y² = 25 intersect.
Expand the second: x² − 14x + 49 + y² = 25
Subtract the first from the second:
(x² − 14x + 49 + y²) − (x² + y²) = 25 − 25
−14x + 49 = 0
x = 7/2
Substitute into x² + y² = 25:
49/4 + y² = 25
y² = 25 − 49/4 = 51/4
y = ±√51/2
Points: (7/2, √51/2) and (7/2, −√51/2)
Key technique: When finding where two circles intersect, subtract one equation from the other. This eliminates the squared terms and gives a linear equation — the equation of the common chord (or radical axis).
The number of intersections between a line and a conic (parabola, circle, ellipse, hyperbola) is at most 2.
Two different conics can intersect in at most 4 points (by Bezout's theorem applied to degree-2 equations).
When one or both curves are given parametrically, substitute the parametric expressions.
Example 9: The curve C₁ has parametric equations x = 2t, y = t². The line L has equation y = 3x − 4. Find where L meets C₁.
Substitute: t² = 3(2t) − 4
t² = 6t − 4
t² − 6t + 4 = 0
t = (6 ± √(36 − 16)) / 2 = (6 ± √20) / 2 = 3 ± √5
When t = 3 + √5: x = 6 + 2√5, y = (3 + √5)² = 14 + 6√5
When t = 3 − √5: x = 6 − 2√5, y = (3 − √5)² = 14 − 6√5
Once you have found intersection points, you may be asked to find the area between the curves.
Example 10: Find the area enclosed between y = x² and y = 2x.
Find intersections: x² = 2x ⟹ x² − 2x = 0 ⟹ x(x − 2) = 0 ⟹ x = 0, 2
Between x = 0 and x = 2, the line y = 2x lies above y = x².
Area = ∫₀² (2x − x²) dx
= [x² − x³/3]₀²
= (4 − 8/3) − 0
= 4/3
Problem: The curve C has equation y = x² − 4x + 7. The line L passes through the point (0, 3) with gradient m.
(a) Show that if L is tangent to C, then m² − 8m + 16 = 0, and hence find m.
(b) Find the coordinates of the point of tangency.
(c) Find the set of values of m for which L does not intersect C.
Solution:
(a) Equation of L: y = mx + 3.
Substitute into C:
mx + 3 = x² − 4x + 7
x² − (4 + m)x + 4 = 0
For tangency: Δ = 0
(4 + m)² − 16 = 0
16 + 8m + m² − 16 = 0
m² + 8m = 0
m(m + 8) = 0
m = 0 or m = −8
Wait — let me recheck. The question states m² − 8m + 16 = 0. Let me redo:
mx + 3 = x² − 4x + 7
x² − (4 + m)x + (7 − 3) = 0
x² − (4 + m)x + 4 = 0
Δ = (4 + m)² − 16 = m² + 8m + 16 − 16 = m² + 8m
Setting Δ = 0: m² + 8m = 0 ⟹ m(m + 8) = 0 ⟹ m = 0 or m = −8.
So L is tangent when m = 0 or m = −8.
(b)
When m = 0: x² − 4x + 4 = 0 ⟹ (x − 2)² = 0 ⟹ x = 2, y = 3. Point: (2, 3).
When m = −8: x² + 4x + 4 = 0 ⟹ (x + 2)² = 0 ⟹ x = −2, y = 19. Point: (−2, 19).
(c) L does not intersect C when Δ < 0:
m² + 8m < 0
m(m + 8) < 0
−8 < m < 0
Exam Tip: When solving intersection problems, always simplify the resulting equation fully before solving. If the question asks you to "show that" a particular equation arises, work carefully to match the given form. For tangency questions, remember that "tangent" means the discriminant is zero. When finding areas between curves, always identify which curve is on top in the given interval.
AQA 7357 specification, Paper 1 — Pure Mathematics, Section C: Coordinate Geometry. AQA requires students to "find the intersection of a line with a curve, including a circle and a parabola; understand and use the relationship between the discriminant of a quadratic and the number of intersection points." Although Section C is the named home, intersection problems thread through the whole specification: Section B (algebra and functions) supplies the simultaneous-equation machinery, Section D (sequences) provides parametric back-substitution patterns, and Section G (differentiation) needs intersection logic to confirm tangency at a stationary point. The AQA formula booklet does not list a "discriminant test for tangency" — students must reconstruct it from b2−4ac.
Question (8 marks):
The line ℓ has equation y=2x+k and the circle C has equation x2+y2=5.
(a) Show that the x-coordinates of the intersection points satisfy 5x2+4kx+(k2−5)=0. (3)
(b) Find the value of k>0 for which ℓ is tangent to C, and state the coordinates of the point of tangency. (5)
Solution with mark scheme:
(a) Step 1 — substitute the line into the circle.
Substituting y=2x+k into x2+y2=5:
x2+(2x+k)2=5
M1 — correct substitution direction (line into circle, since the line is given explicitly in y). Substituting the other way is much messier and rarely earns marks. Examiners look for the cleaner substitution.
Step 2 — expand and collect.
x2+4x2+4kx+k2=5 5x2+4kx+(k2−5)=0
M1 — correct expansion of (2x+k)2=4x2+4kx+k2. Common error: writing (2x+k)2=4x2+k2, missing the 4kx cross term. That single slip costs M1 and propagates through every later mark.
A1 — printed result obtained, with (k2−5) correctly bracketed as the constant term.
(b) Step 1 — apply the discriminant test for tangency.
Tangency means the quadratic has exactly one repeated root, so b2−4ac=0 with a=5, b=4k, c=k2−5:
(4k)2−4⋅5⋅(k2−5)=0
M1 — correctly identifying that tangency ⇔ discriminant =0, with the coefficients read off correctly.
Step 2 — solve for k.
16k2−20(k2−5)=0 16k2−20k2+100=0 −4k2+100=0 k2=25⟹k=±5
M1 — correct algebraic manipulation through to k2=25.
A1 — k=5 selected (the positive root demanded by the question stem). Stating k=±5 without naming k=5 explicitly costs the A1.
Step 3 — find the point of tangency.
With k=5 the quadratic becomes 5x2+20x+20=0, i.e. x2+4x+4=0, so (x+2)2=0 and x=−2. Then y=2(−2)+5=1.
M1 — back-substituting k=5 to obtain the repeated root and the corresponding y.
A1 — point of tangency (−2,1), with both coordinates verified against the circle ((−2)2+12=5 ✓).
Total: 8 marks (M5 A3, split as shown).
Question (6 marks): The curve y=x2−3x+4 and the line y=mx+1 meet at exactly one point.
(a) Show that m satisfies (m+3)2=12. (3)
(b) Hence find both possible values of m in surd form. (3)
Mark scheme decomposition by AO:
(a)
(b)
Total: 6 marks split AO1 = 5, AO2 = 1. AQA uses tangency questions to test AO1 (the discriminant procedure) with a single AO2 mark for recognising that "exactly one point" translates algebraically to "discriminant zero". Candidates who write "set the equations equal and solve" without invoking the discriminant lose AO2 and usually one of the M1s.
Connects to:
Section B — Algebra (simultaneous equations): the substitution method for solving line-curve intersections is exactly the elimination technique used in linear simultaneous equations, generalised to a non-linear curve. The same algebraic discipline (substitute the simpler equation into the harder) carries over directly.
Section B — Quadratic discriminant: the "three cases" of intersection (two points, one point, no points) map one-to-one onto the three cases of b2−4ac (positive, zero, negative). Mastery of the discriminant in Section B is literally mastery of intersection geometry — it is the same theorem dressed up.
Section C — Circles: the centre-radius form (x−a)2+(y−b)2=r2 is the standard input for line-circle problems. Completing the square to recover centre and radius is a Section B technique deployed in a Section C context.
Section D — Parametric equations: for parametric curves x=f(t), y=g(t), intersection with a line y=mx+c becomes g(t)=mf(t)+c — a single equation in t. The discriminant test still tells you about tangency, applied to that t-equation.
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