In DSE Paper 1 Section B, you will often encounter systems that combine a linear equation with a quadratic equation. Geometrically, this means finding the intersection points of a line and a parabola (or a circle). This article teaches you the substitution method for solving such systems: solve the linear equation for one variable, substitute into the quadratic, and solve the resulting quadratic equation. You will learn how to handle two intersections, one intersection (tangent), and no intersection cases. Worked examples, exam techniques, and plenty of practice are included.
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Cross-Topic: Simultaneous Equations + Quadratic Equations
Exercise
Solve each system or find the required values.
- \(\begin{cases} y = x - 1 \\ y = x^2 - 3x + 2 \end{cases}\)
- \(\begin{cases} y = 2x + 3 \\ y = x^2 + x - 1 \end{cases}\)
- \(\begin{cases} y = -x + 2 \\ y = x^2 + 2x + 1 \end{cases}\)
- Determine the number of intersections of line \(y = 3x-2\) and parabola \(y = x^2-2x+4\).
- Determine the number of intersections of line \(y = x+5\) and parabola \(y = x^2-3x+7\).
- Find \(k\) such that line \(y = kx-3\) is tangent to parabola \(y = x^2-4x+5\).
- Find the intersection points of the line \(y = x+3\) and the circle \(x^2 + y^2 = 25\).
- Find the value of \(m\) such that the line \(y = mx+1\) intersects the parabola \(y = x^2-2x+2\) at exactly one point.
- \(\begin{cases} y = 2x - 5 \\ y = x^2 - 6x + 10 \end{cases}\)
- If the line \(y = x + c\) is tangent to the parabola \(y = x^2 - 2x + 4\), find \(c\).
Answer Key
1. \((1,0)\) and \((3,2)\)
2. \((-1,1)\) and \((4,11)\)
3. \((\frac{-3+\sqrt{5}}{2}, \frac{7-\sqrt{5}}{2})\) and \((\frac{-3-\sqrt{5}}{2}, \frac{7+\sqrt{5}}{2})\)
4. 0 (discriminant \(<0\))
5. 2 (discriminant \(>0\))
6. \(k = 2 \pm 2\sqrt{2}\)
7. \((1,4)\) and \((-4,-1)\)
8. \(m = -2 \pm \sqrt{2}\)
9. \((3,1)\) and \((5,5)\)
10. \(c = 4\) (since tangent when discriminant zero: \(x+c = x^2-2x+4 \Rightarrow x^2-3x+(4-c)=0\), disc = \(9-4(4-c)=0 \Rightarrow 9-16+4c=0 \Rightarrow 4c=7 \Rightarrow c=\frac{7}{4}\). Wait: let's recalc. Actually: \(x^2-2x+4 = x+c \Rightarrow x^2-3x+(4-c)=0\). Discriminant: \(9 - 4(4-c) = 9 - 16 + 4c = 4c - 7 = 0 \Rightarrow c = 7/4\). So answer: \(c = \frac{7}{4}\).