Today's lesson
A straight line can cut a parabola twice, just touch it once, or miss it completely. We find the meeting points by solving the two equations simultaneously (substitution → a quadratic), and the discriminant tells us how many meeting points there are.
Learning intentions
- Understand that a line can meet a parabola at 0, 1 or 2 points
- Find the points of intersection using substitution
- Use the discriminant to find the number of intersection points
- State each intersection as a coordinate pair $(x,y)$
Part 1 — Three possibilities (~8 min)
Solving a line and a parabola together gives an equation $ax^2+bx+c=0$. Its number of solutions is the number of points of intersection.
Key ideas
- A line that cuts the curve twice is a secant; a line that touches once is a tangent.
- Method of substitution: substitute one equation into the other, rearrange to $ax^2+bx+c=0$, solve (factorise or quadratic formula), then substitute back to find each $y$.
- The discriminant $b^2-4ac$ of that quadratic gives the number of intersection points:
$<0\Rightarrow$ zero (miss); $=0\Rightarrow$ one (tangent); $>0\Rightarrow$ two (secant). - Always give the answer as coordinate pairs: when $x=\dots$, then $y=\dots$.
Part 2 — Solving by substitution (Example 1, ~14 min)
By substitution: $x^2=2x \Rightarrow x^2-2x=0 \Rightarrow x(x-2)=0$, so $x=0$ or $x=2$.
When $x=0$, $y=2(0)=0$. When $x=2$, $y=2(2)=4$.
Points of intersection: $(0,0)$ and $(2,4)$. Two points, so the line is a secant.
1. Substitute one equation into the other.
2. Rearrange to $ax^2+bx+c=0$.
3. Solve for $x$ (factorise or quadratic formula).
4. Substitute each $x$ back to find $y$.
5. State each point as $(x,y)$.
Now you try: Find the intersection of $y=x^2$ and $y=4x$. Answer: $x^2=4x \Rightarrow x(x-4)=0$, so $x=0,4$, giving $(0,0)$ and $(4,16)$.
Part 3 — Discriminant & the quadratic formula (Example 2, ~12 min)
a) $-4x^2-x+6=3x+7 \Rightarrow 0=4x^2+4x+1 \Rightarrow (2x+1)^2=0$, so $x=-\tfrac12$ (one solution — a tangent). Then $y=3\left(-\tfrac12\right)+7=\tfrac{11}{2}$. Point of intersection $\left(-\tfrac12,\tfrac{11}{2}\right)$. The discriminant $4^2-4(4)(1)=0$ confirms one point.
b) $x^2+5x-5=2x \Rightarrow x^2+3x-5=0$. Quadratic formula: $x=\dfrac{-3\pm\sqrt{9+20}}{2}=\dfrac{-3\pm\sqrt{29}}{2}=1.19$ or $-4.19$ (2 d.p.). Then $y=2x$, giving $y=2.39$ and $y=-8.39$. Points $(1.19,2.39)$ and $(-4.19,-8.39)$.
Now you try: Solve $y=x^2+2x-1$ and $y=3x$ (2 d.p.). Answer: $x^2-x-1=0$, so $x=1.62$ or $-0.62$; points $(1.62,4.85)$ and $(-0.62,-1.85)$.
📺 Walkthrough: solving $y=x^2$ and $y=2x$ by substitution to find both meeting points, $(0,0)$ and $(2,4)$.
Practice 3.1 — find the points of intersection (or state "none").
- $y=x^2-3x$ and $y=4$
- $y=x^2-x$ and $y=2$
- $y=x^2+2x+4$ and $y=-2$
- $y=x^2+1$ and $2x-3y=-4$
b) $x^2-x-2=0 \Rightarrow x=2,-1$: $(2,2)$ and $(-1,2)$
c) $x^2+2x+6=0$, $\Delta=-20<0$: none
d) $3x^2-2x-1=0 \Rightarrow x=1,-\tfrac13$: $(1,2)$ and $\left(-\tfrac13,\tfrac{10}{9}\right)$
Part 4 — Quick quiz (5 min)
Pick the correct answer for each, then click Mark.
Q1. A line that cuts a parabola at exactly two points is called a:
Q2. After substituting, you get $ax^2+bx+c=0$ with $\Delta=0$. The line is a:
Q3. Solving $y=x^2$ and $y=2x$ gives $x(x-2)=0$. The intersection points are:
Q4. If the resulting quadratic has $\Delta<0$, the line and parabola meet at:
Q5. The first step to find where a line meets a parabola is to:
Working program — Cambridge Ex 7G (p638)
After the quiz, open Cambridge Chapter 7 (page 638) and complete the following:
| Set work | Extension |
|---|---|
| Questions 1–3 (½), 4, 5 (½), 6, 7 (½), 9, 10 | Questions 8, 11, 12 |
"(½)" means do every second part. Always give each intersection as a coordinate pair $(x,y)$.
Exit ticket — write in your book
- How many points can a line and a parabola have in common?
- What method do we use to find those points, and what kind of equation does it give?
- What does $\Delta=0$ tell you about the line (what is the line called)?