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Quadratics

Quadratics

Where a line meets a curve: intersections, tangents and the discriminant

Intersection of a Line and a Curve AS Level 19 min read

On this page
  1. What you need to be able to do
  2. What you should already know
  3. What "where they meet" means
  4. Solving a linear and a quadratic equation together
  5. Intersect, touch or miss: the discriminant again
  6. Case by case
  7. Tangents: the calculus route
  8. Key facts at a glance
  9. How this is examined
  10. Mark-scheme focus
  11. Worked examples
  12. Summary
  13. Can I do this?
  14. Quiz

What you need to be able to do

Syllabus 1.1 and 1.3
  • 1.1Solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic.
  • 1.3Use the link between points of intersection and the roots of an equation: find the set of values of \(k\) for which a line intersects, touches or does not meet a curve.
  • 1.3Find a tangent (the value of the constant and the point of contact), and prove that a line meets a curve for all values of a constant.

Before you start: The discriminant and Quadratic inequalities. The same method works for lines and circles: see Coordinate geometry.

Part 1Learn the topic

What it means for a line to meet a curve, how to find the meeting points, and how the discriminant counts them, with our own examples. Past-paper questions come in Part 2.

What you should already know

Simultaneous linear equations

Two equations, two unknowns: make one letter the subject of one equation and substitute into the other. \(y=x+1\) and \(2x+y=7\): \(2x+x+1=7\Rightarrow x=2,\ y=3\).

Quick check: solve \(y=3x-1,\ x+y=7\)

\(4x-1=7\Rightarrow x=2,\ y=5\)

Solving a quadratic

Factorise, complete the square or use the formula: Solving quadratic equations.

Quick check: solve \(x^2-x-12=0\)

\((x-4)(x+3)=0\): \(x=4,\ -3\)

The discriminant

\(b^2-4ac>0\): two roots; \(=0\): one repeated root; \(<0\): none (previous pages).

Quick check: how many roots has \(x^2+2x+5=0\)?

\(4-20<0\): none

A point lies on a graph

A point is on a graph exactly when its coordinates satisfy the equation. \((2,\,7)\) is on \(y=x^2+3\) because \(2^2+3=7\).

Quick check: is \((-1,\,4)\) on \(y=x^2+3\)?

\((-1)^2+3=4\): yes

What "where they meet" means

A meeting point of a line and a curve lies on both graphs, so its coordinates satisfy both equations at once. Finding the meeting points is therefore the same as solving the two equations simultaneously.

When you substitute the line into the curve you get one equation in \(x\). Each root of that equation is the \(x\)-coordinate of a meeting point. That single fact explains everything on this page: two roots means two meeting points, a repeated root means the line just touches (a tangent), no real roots means the line misses the curve.

Solving a linear and a quadratic equation together

The points where a line and a curve meet satisfy both equations. Make one letter the subject of the linear equation, substitute into the quadratic one, and you get a quadratic in a single letter.

  1. Rearrange the linear equation for the letter that is easiest to substitute (no fractions if possible).
  2. Substitute into the other equation, with brackets round the whole expression: \(x(3-x)+(3-x)^2\), not \(x\cdot3-x+3-x^2\).
  3. Expand and collect to a 3-term quadratic \(=0\).
  4. Solve it.
  5. Find the other coordinate from the linear equation (quicker and safer than the quadratic), and pair the values: \((x_1,\ y_1)\) and \((x_2,\ y_2)\).

Example. Find where the line \(y=2x-2\) meets the curve \(y=x^2-3x+4\).

  1. The line already gives \(y\). Substitute: \(x^2-3x+4=2x-2\)
  2. Collect: \(x^2-5x+6=0\Rightarrow(x-2)(x-3)=0\), so \(x=2\) or \(x=3\)
  3. From the line: \(x=2\Rightarrow y=2\); \(x=3\Rightarrow y=4\)
  4. Meeting points \((2,\,2)\) and \((3,\,4)\). Check in the curve: \(9-9+4=4\) ✓

Example where the curve is not \(y=\dots\): solve \(x+y=3\) and \(x^2+xy+y^2=7\). From the line \(y=3-x\): \(x^2+x(3-x)+(3-x)^2=7\Rightarrow x^2-3x+2=0\), so \(x=1,\ y=2\) or \(x=2,\ y=1\).

Which letter to eliminate? The one the line gives most simply. If the curve is written \(y=\dots\), eliminate \(y\): substituting \(x\) instead usually creates fractions and a much messier quadratic.

Try it yourself
  1. Find where \(y=x+3\) meets \(y=x^2+1\).
    Answer\(x^2-x-2=0\): \(x=2,\ -1\); points \((2,\,5)\) and \((-1,\,2)\)
  2. Solve \(y=2x+1\) and \(x^2+y^2=2\).
    Answer\(5x^2+4x-1=0\Rightarrow(5x-1)(x+1)=0\): \(\left(\tfrac15,\tfrac75\right)\) and \((-1,-1)\)

Intersect, touch or miss: the discriminant again

xyOy = x²y = 2x + 2: cuts twicey = 2x − 1: touches (tangent)y = 2x − 3.5: misses
Lines with the same gradient: two points, one point (tangent), no points.
  1. Equate the curve and the line and rearrange to \(ax^2+bx+c=0\). Each root is the \(x\)-coordinate of a meeting point.
  2. Two distinct points (cuts the curve): \(b^2-4ac>0\).
  3. One point, a tangent (touches): \(b^2-4ac=0\). The repeated root is the point of contact.
  4. Does not meet: \(b^2-4ac<0\).
  5. Meets / intersects (at least once): \(b^2-4ac\ge0\).

So every "set of values of \(k\)" question is the same three-stage job: form the quadratic (equate and collect, with \(k\) in the coefficients) → apply the condition → solve the inequality in \(k\). The first stage is where most slips happen: move every term to one side and keep the signs.

Case by case

1 Two distinct points: set of k

Line \(y=2x+k\), curve \(y=x^2+kx+5\). Find the values of \(k\) for which they meet at two distinct points.

Askedset of kKnowk in both equationsRouteequate, collect, D > 0, outside

  1. \(x^2+(k-2)x+(5-k)=0\)
  2. \((k-2)^2-4(5-k)>0\Rightarrow k^2-4k+4-20+4k>0\Rightarrow k^2-16>0\)
  3. \((k+4)(k-4)>0\Rightarrow\mathbf{k<-4,\ k>4}\)

2 Does not meet: set of k

Curve \(y=kx^2+3x-k\), line \(y=kx-3\). Find the values of \(k\) for which they do not intersect.

Askedset of kKnowk in a, b and cRouteD < 0, between

  1. \(kx^2+(3-k)x+(3-k)=0\)
  2. \((3-k)^2-4k(3-k)<0\Rightarrow(3-k)(3-5k)<0\)
  3. \(\mathbf{\tfrac35<k<3}\)

Factor out (3 − k) instead of expanding: (3 − k)[(3 − k) − 4k]. Less algebra, fewer slips.

3 Tangent: the constant and the point

\(y=mx-3\) is a tangent to \(y=x^2+2x+1\). Find \(m\) and the points of contact.

Askedm and (x, y)Knowtangent: one repeated rootRouteD = 0 gives m; put each m back to get x, then y

  1. \(x^2+(2-m)x+4=0\); \((2-m)^2-16=0\Rightarrow2-m=\pm4\)
  2. \(m=6\): \(x^2-4x+4=0\Rightarrow x=2,\ y=9\)
  3. \(m=-2\): \(x^2+4x+4=0\Rightarrow x=-2,\ y=1\)

Each value of m gives its own tangent and its own point of contact. Pair them up.

4 Meet for all values of k

Show that \(y=x+k\) and \(y=x^2+kx\) meet for all values of \(k\).

Askeda proofKnowthe discriminant will be an expression in kRouteshow D ≥ 0 whatever k is, and say so

  1. \(x^2+(k-1)x-k=0\)
  2. \(b^2-4ac=(k-1)^2+4k=k^2+2k+1=(k+1)^2\)
  3. \((k+1)^2\ge0\) for all \(k\), so the line and curve meet for all values of \(k\)

Finding the perfect square is not the end: say what it means. A square is never negative, so D ≥ 0, so the line always meets the curve.

5 Two distinct points for all values of m

Show that \(y=x^2+m(x+1)\) and \(y=x+3\) meet at two distinct points for all \(m\).

Askeda proofKnowD is a quadratic in m that doesn't factoriseRoutecomplete the square on D, show it is always positive

  1. \(x^2+(m-1)x+(m-3)=0\)
  2. \(b^2-4ac=(m-1)^2-4(m-3)=m^2-6m+13=(m-3)^2+4\)
  3. \((m-3)^2+4>0\) for all \(m\), so \(b^2-4ac>0\): two distinct points for all values of m

"For all m" means you cannot pick a value of m or solve D = 0. Complete the square on D to show it can never be zero or negative.

Tangents: the calculus route

For a tangent to \(y=f(x)\) there is a second method (on Paper 1 once you know differentiation): the gradient of the curve at the point of contact equals the gradient of the line. Both methods are correct.

Both methods on one question: \(y=mx+1\) is a tangent to \(y=x^2-2x+5\). Find \(m\) and the points of contact.

Method 1: discriminant = 0

  1. \(x^2-2x+5=mx+1\Rightarrow x^2-(2+m)x+4=0\)
  2. Tangent: \((2+m)^2-4(1)(4)=0\Rightarrow(2+m)^2=16\)
  3. \(m=2\) or \(m=-6\)
  4. Substitute each m back: \(x^2-4x+4=0\Rightarrow x=2\); \(x^2+4x+4=0\Rightarrow x=-2\). Then \(y\) from the line: \((2,5)\), \((-2,13)\)

Works for any curve whose equation gives a quadratic, including circles and xy = k.

Method 2: gradients equal

  1. \(\dfrac{dy}{dx}=2x-2\), and the line's gradient is \(m\): \(2x-2=m\)
  2. At the contact point the line and curve also meet: \(x^2-2x+5=(2x-2)x+1\)
  3. \(x^2-2x+5=2x^2-2x+1\Rightarrow x^2=4\Rightarrow x=\pm2\)
  4. \(x=2\): \(m=2,\ y=5\); \(x=-2\): \(m=-6,\ y=13\)

Often the more direct route: the quadratic you solve (here x² = 4) is simpler.

Try it yourself
  1. Find \(k\) for which \(y=x+k\) is a tangent to \(y=x^2-3x+5\), and the point of contact.
    Answer\(x^2-4x+5-k=0\); \(16-4(5-k)=0\Rightarrow k=1\); \(x=2,\ y=3\)
  2. Find the set of values of \(k\) for which \(y=kx-1\) does not meet \(y=x^2+3\).
    Answer\(x^2-kx+4=0\); \(k^2-16<0\): \(-4<k<4\)

Key facts at a glance

  • Meeting points satisfy both equations: substitute the line into the curve.
  • Each root of the resulting quadratic is the \(x\)-coordinate of a meeting point; find \(y\) from the line.
  • \(b^2-4ac>0\) cuts twice; \(=0\) tangent (touches); \(<0\) misses; \(\ge0\) meets.
  • Tangent questions: \(D=0\) gives the constant, the repeated root gives the point.
  • "For all values of \(k\)": show \(D\) is a square (\(\ge0\)) or a completed square plus a positive (\(>0\)), and say so.
Part 2Past-paper questions, step by step

How lines and curves are examined, what the mark scheme wants, then real Paper 1 questions worked through slowly.

How this is examined

Some 14 questions since 2020 ask for the set of \(k\) for which a line meets, touches or misses a curve (3-7 marks), and about 6 more for a tangent's constant and point of contact. The marks: M1 for substituting to form the quadratic, DM1 for collecting it to \(=0\) and using \(b^2-4ac\), A1 A1 for the values. The November 2021 report: "sign errors were common, leading to the wrong quadratic equation".

"Show that ... for all values of \(k\)": the March 2023 report found the perfect square often appeared, but "the correct interpretation of this was rarely seen"; the November 2020 report praised candidates who "realised the need to use completing the square ... and were then able to explain". The June 2021 report noted that the differentiation route to a tangent "resulted in a more direct route to the answers".

Which letter should you eliminate? The November 2020 report on 9709/11/O/N/20 Q1: "All successful attempts at this question involved the elimination of \(y\)... those who chose to eliminate \(x\) often made little progress." If the curve is \(y=\dots\), eliminate \(y\). Substitute into the curve the letter that the line gives most simply, usually y = mx + c.

Mark-scheme focus

Show that the line and curve meet for all values of k

9709/12/F/M/23 Q1Paper 1 · Pure Mathematics 14 marksFull size
A line has equation y = 3x − 2k and a curve has equation y = x² − kx + 2, where k is a constant. Show that the line and the curve meet for all values of k. [4]
Credited
  • 3-term quadratic M1; \(b^2-4ac=(k+3)^2-4(2+2k)\) DM1
  • \(=(k-1)^2\) A1
  • "\(\ge0\). Hence will meet for all values of k" - a clear conclusion A1
Not credited
  • \((k-1)^2\) with no comment
  • "\(>0\)" (it equals 0 when \(k=1\): there they touch)
  • checking a few values of k
Also accepted

calculus to show the minimum of \(k^2-2k+1\) is 0, or a sketch of \(f(k)=k^2-2k+1\).

From: 9709/12/F/M/23 Q1 and the March 2023 report.

A tangent: find the constant AND the point of contact

9709/11/M/J/23 Q5Paper 1 · Pure Mathematics 15 marksFull size
The line with equation y = kx − k, where k is a positive constant, is a tangent to the curve with equation y = −1/(2x). Find, in either order, the value of k and the coordinates of the point where the tangent meets the curve. [5]
Credited
  • \(kx-k=-\dfrac1{2x}\Rightarrow2kx^2-2kx+1=0\), all terms on one side *M1
  • \(b^2-4ac=4k^2-8k=0\), the discriminant used on its own DM1
  • \(k=2\) only: \(k=0\) is rejected because \(k\) is positive A1
  • \(4x^2-4x+1=0\Rightarrow x=\tfrac12\) B1; \(y=-1\) B1
Not credited
  • \(k=0\) or \(k=2\) left as the answer
  • stopping at \(k\): the point of contact is 2 of the 5 marks
  • \(b^2-4ac\) only inside the quadratic formula (with \(x\) still present: DM0)
Why

"Tangent" gives one equation for k, but the question also wants where it touches. Put k back into the quadratic: its repeated root is the x-coordinate.

Also accepted

Calculus: \(\dfrac{dy}{dx}=\dfrac1{2x^2}=k\), then solve with the line and curve together.

From: 9709/11/M/J/23 Q5.

Worked examples

Example 1 · Intersection of a line and a curve (Paper 1, 4 marks) ★ Easy
9709/12/M/J/25 Q2Paper 1 · Pure Mathematics 14 marksFull size
Find the coordinates of the points of intersection of the curve and the line with equations 2xy + 5y² = 24 and 2x + y + 4 = 0. [4]
1 Read it with me what is actually happening?

The curve isn't in \(y=\dots\) form, so look at the line: \(y=-2x-4\) has no fractions. Substitute that into the curve and you get a quadratic in \(x\).

  • coordinatespairs (x, y): find both coordinates of each point
  • 2x + y + 4 = 0\(y=-2x-4\)
2 What is it asking for? the target, part by part
  • two points (x, y)4 marks
3 What do we know? everything the question gives us
  • \(2xy+5y^2=24\)curve
  • \(2x+y+4=0\)line
4 Plan the route work backwards from what we want
FindMaths ideaAlready have itWork out firstTrap
  • Findboth points
    • Use\(y=-2x-4\) into the curve
      • Then\(16x^2+72x+56=0\), ÷8: \(2x^2+9x+7=0\)
    • Backy from the line
Expand 5(−2x − 4)².

\(5(4x^2+16x+16)=20x^2+80x+80\).

Why find y from the line, not the curve?

The line gives one y for each x; the curve might give two, and it's more work.

5 Now solve it have a go on paper first, then open the worked solution
  1. \(2x(-2x-4)+5(-2x-4)^2=24\) *M1
  2. \(-4x^2-8x+20x^2+80x+80=24\Rightarrow16x^2+72x+56=0\Rightarrow2x^2+9x+7=0\) DM1
  3. \((2x+7)(x+1)=0\Rightarrow x=-1,\ -\tfrac72\) B1
  4. \(y=-2x-4\): \(\mathbf{(-1,\ -2)}\) and \(\mathbf{\left(-\tfrac72,\ 3\right)}\) B1
6 Check it does the answer make sense?

\((-1,-2)\) in the curve: \(2(-1)(-2)+5(4)=4+20=24\) ✓. \((-\tfrac72,3)\): \(-21+45=24\) ✓.

Example 2 · Coordinates, then a set of k (Paper 1, 9 marks) ★★ Medium
9709/11/M/J/25 Q6Paper 1 · Pure Mathematics 19 marksFull size
The equation of a curve is 2x² − kxy + 2 = 0 and the equation of a line is y = px + 3, where k and p are constants. (a) Given that k = 2 and p = 11, find the coordinates of the points of intersection of the curve and the line. [4] (b) Given instead that p = 4, find the set of values of k for which the curve and the line do not intersect. [5]
1 Read it with me what is actually happening?

Same pair of equations, two different jobs. In (a) all the constants are known: solve. In (b) \(k\) is unknown and we want no intersection: discriminant \(<0\). "Given instead" means (a)'s numbers do not carry over.

  • Given insteadstart (b) afresh with p = 4; k is unknown
  • do not intersect\(b^2-4ac<0\)
  • −kxyafter substituting \(y\), this term has \(x^2\) in it: it joins the \(x^2\) coefficient
2 What is it asking for? the target, part by part
  • (a)both points4 marks
  • (b)set of k5 marks
3 What do we know? everything the question gives us
  • \(2x^2-kxy+2=0\)curve
  • \(y=px+3\)line
  • in (b) the \(x^2\) coefficient is \(2-4k\): it has two parts
4 Plan the route work backwards from what we want
FindMaths ideaAlready have itWork out firstTrap
  • (a)points
    • Use\(2x^2-2x(11x+3)+2=0\Rightarrow10x^2+3x-1=0\)
  • (b)set of k
    • Form\((2-4k)x^2-3kx+2=0\)
    • Use\(9k^2-8(2-4k)<0\)
    • Regionbetween
In (b), what is a, the coefficient of x²?

\(2-4k\): from \(2x^2\) and from \(-kx\cdot4x\).

Solve 9k² + 32k − 16 = 0.

\((9k-4)(k+4)=0\Rightarrow k=\tfrac49\) or \(-4\).

5 Now solve it have a go on paper first, then open the worked solution
  1. (a) \(2x^2-2x(11x+3)+2=0\Rightarrow-20x^2-6x+2=0\) *M1 \(\Rightarrow(5x-1)(2x+1)=0\) DM1
  2. \(\left(\tfrac15,\ \tfrac{26}{5}\right)\) and \(\left(-\tfrac12,\ -\tfrac52\right)\) A1 A1
  3. (b) \(2x^2-kx(4x+3)+2=0\Rightarrow(2-4k)x^2-3kx+2=0\) *M1 A1
  4. \(b^2-4ac=9k^2-8(2-4k)<0\) DM1 \(\Rightarrow9k^2+32k-16<0\Rightarrow(9k-4)(k+4)<0\) M1
  5. \(\mathbf{-4<k<\tfrac49}\) A1
6 Check it does the answer make sense?

(a): \(y=11(\tfrac15)+3=\tfrac{26}5\) ✓. (b): \(k=0\) (inside) gives \(2x^2+2=0\), no real roots ✓.

Example 3 · Tangent: two values of c and two points (Paper 1, 7 marks) ★★ Medium
9709/13/O/N/23 Q6Paper 1 · Pure Mathematics 17 marksFull size
A line has equation y = 6x − c and a curve has equation y = cx² + 2x − 3, where c is a constant. The line is a tangent to the curve at point P. Find the possible values of c and the corresponding coordinates of P. [7]
1 Read it with me what is actually happening?

The constant \(c\) appears in both equations. Tangent means the equation for the meeting points has one repeated root. That gives an equation in \(c\) with two answers - and each answer has its own point \(P\).

  • tangent\(b^2-4ac=0\)
  • corresponding coordinatesone P for each value of c, paired
2 What is it asking for? the target, part by part
  • values of c and P for each7 marks
3 What do we know? everything the question gives us
  • \(y=6x-c\)line
  • \(y=cx^2+2x-3\)curve
4 Plan the route work backwards from what we want
FindMaths ideaAlready have itWork out firstTrap
  • Findc and P
    • Form\(cx^2-4x+(c-3)=0\)
    • Use\(16-4c(c-3)=0\Rightarrow c^2-3c-4=0\)
    • Thenput each c back, solve the (perfect square) quadratic, y from the line
What are the two values of c?

\((c-4)(c+1)=0\): \(c=4\) or \(c=-1\).

Why is the quadratic a perfect square when you put c back?

D = 0 by construction, so it has one repeated root.

5 Now solve it have a go on paper first, then open the worked solution
  1. \(cx^2+2x-3=6x-c\Rightarrow cx^2-4x+(c-3)=0\) B1
  2. \(16-4c(c-3)=0\) *M1 \(\Rightarrow c^2-3c-4=0\Rightarrow c=4,\ -1\) A1
  3. \(c=4\): \(4x^2-4x+1=0\Rightarrow(2x-1)^2=0\), \(x=\tfrac12,\ y=-1\) DM1 A1
  4. \(c=-1\): \(-x^2-4x-4=0\Rightarrow(x+2)^2=0\), \(x=-2,\ y=-11\) DM1 A1

Alternative on the scheme: \(\tfrac{dy}{dx}=2cx+2=6\Rightarrow c=\tfrac2x\); substitute into the meeting equation to get \(2x^2+3x-2=0\Rightarrow x=\tfrac12\) or \(-2\).

6 Check it does the answer make sense?

\(c=4\): curve at \(x=\tfrac12\): \(1+1-3=-1\) ✓ matches the line \(3-4=-1\).

Example 4 · A line touching an ellipse-shaped curve (Paper 1, 6 marks) ★★ Medium
9709/12/F/M/24 Q7Paper 1 · Pure Mathematics 16 marksFull size
The straight line y = x + 5 meets the curve 2x² + 3y² = k at a single point P. (a) Find the value of the constant k. [4] (b) Find the coordinates of P. [2]
1 Read it with me what is actually happening?

"A single point" for a line and a curve means the line touches it: a repeated root. Substitute the line into the curve, set the discriminant to zero, find \(k\), then the point.

  • single point\(b^2-4ac=0\)
  • 2x² + 3y² = ka curve (an ellipse); substitute \(y=x+5\)
2 What is it asking for? the target, part by part
  • (a)k4 marks
  • (b)P2 marks
3 What do we know? everything the question gives us
  • \(y=x+5\)line
  • \(2x^2+3y^2=k\)curve
4 Plan the route work backwards from what we want
FindMaths ideaAlready have itWork out firstTrap
  • (a)k
    • Form\(5x^2+30x+75-k=0\)
    • Use\(900-20(75-k)=0\)
  • (b)P
    • Use\(k=30\): \(5x^2+30x+45=0\Rightarrow(x+3)^2=0\)
Expand 3(x + 5)².

\(3x^2+30x+75\).

What is x at the repeated root?

\(x=-\tfrac{b}{2a}=-\tfrac{30}{10}=-3\).

5 Now solve it have a go on paper first, then open the worked solution
  1. \(2x^2+3(x+5)^2=k\Rightarrow5x^2+30x+75-k=0\) *M1 A1
  2. \(900-20(75-k)=0\) DM1 \(\Rightarrow k=30\) A1
  3. (b) \(5x^2+30x+45=0\Rightarrow x=-3\) M1, \(y=2\): \(\mathbf{P(-3,\ 2)}\) A1
6 Check it does the answer make sense?

\(2(9)+3(4)=30=k\) ✓.

Summary

  • Substitute the line into the curve (usually eliminate y), collect to ax² + bx + c = 0.
  • Two points: D > 0; tangent: D = 0; misses: D < 0; meets: D ≥ 0.
  • Tangent questions: D = 0 for the constant, then put each value back for its point; or use dy/dx = gradient of the line.
  • "For all values of k": write D as a perfect square or a completed square and conclude in words.
If you see…do this
"points of intersection"substitute, solve, pair the coordinates
"tangent", "touches", "single point"D = 0 (then put the constant back for the point)
"does not meet / intersect"D < 0 then an inequality
"two distinct points"D > 0 then an inequality
"meet for all values of k"D = (…)² ≥ 0 or (…)² + positive > 0, and say so
"given instead"start again: earlier answers do not apply

Can I do this?

Anything unticked: revise that section first. All ticked: take the quiz.

Quiz

The multiple-choice and guided questions need JavaScript. The concept checks and exam practice below work without it.

Concept checks

Answer each one in your head first, then open it.

1A line meets a curve where \(3x^2+bx+c=0\) has a negative discriminant. Can the line and curve still cross somewhere?Show answer

No. The roots of that equation are exactly the \(x\)-coordinates of the meeting points; none exist.

2Why must you put the value of \(k\) back in a tangent question?Show answer

\(b^2-4ac=0\) only tells you \(k\). The point of contact comes from the repeated root \(x=-\tfrac{b}{2a}\) of the equation with that \(k\).

3Is a line that meets a curve exactly once always a tangent?Show answer

For a quadratic curve and a non-vertical line, yes (a repeated root). But a vertical line \(x=a\) meets \(y=x^2\) once without touching: that is why Paper 1 lines are given as \(y=mx+c\).

Exam-style practice

New questions, not used above. Do them on paper, then mark your work against the official mark scheme.

Q19709/11/O/N/21 Q2Paper 1 · Pure Mathematics 15 marksFull size
A curve has equation y = kx² + 2x − k and a line has equation y = kx − 2. Find the set of values of k for which the curve and line do not intersect. [5]
Show the mark scheme
PartAnswer / workingMarksGuidance
\(kx^2+(2-k)x+(2-k)=0\)*M1eliminate y, 3-term quadratic; allow 1 error
\((2-k)^2-4k(2-k)\)DM1\(b^2-4ac\) with their \(a,b,c\)
\(5k^2-12k+4\) or \((2-k)(2-k-4k)\)A1
\((5k-2)(k-2)\)DM1solve the 3-term quadratic in k: factors must expand back
\(\tfrac25<k<2\)A1two separate correct inequalities also accepted
Total: 5 marks
Q29709/12/O/N/20 Q3Paper 1 · Pure Mathematics 15 marksFull size
The equation of a curve is y = 2x² + m(2x + 1) and a line is y = 6x + 4. Show that, for all values of m, the line intersects the curve at two distinct points. [5]
Show the mark scheme
PartAnswer / workingMarksGuidance
\(2x^2+(2m-6)x+(m-4)=0\)*M1y eliminated, all terms on one side
\((2m-6)^2-8(m-4)\)DM1discriminant with their coefficients
\(4m^2-32m+68\) (or \(m^2-8m+17\))A1
\(4[(m-4)^2+1]\), or the minimum point \((4,1)\)DM1any valid method to show the sign
always \(>0\), so two distinct points for all values of mA1clear reasoning and conclusion, no wrong working
Total: 5 marks

Examiner tip The last mark is the sentence. A completed square with no conclusion, or a few trial values of m, does not get it.

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