Intersection of Lines And Parabolas
Every question with a fully worked solution.
- Intersection of Lines And Parabolas - Video - Intersection of parabola and line Watch
Theory
When one equation is a curve, substitute the linear equation into the non-linear one. The result is a quadratic, so there can be two, one or no solutions. This Year 10A maths page covers lines with parabolas, hyperbolas and circles, two parabolas, exact answers with surds, tangents, and finding the constant that makes a line touch a curve.
A pair of simultaneous equations is non-linear when at least one of them is not a straight line, such as a parabola \(y = ax^2 + bx + c\), a hyperbola \(y = \dfrac{k}{x}\) (or \(xy = k\)), or a circle \(x^2 + y^2 = r^2\).
Each solution is a point of intersection of the two graphs, so a solution is a pair \((x, y)\). Pair each \(x\) with its own \(y\).
A line that meets a curve at exactly one point without crossing it is a tangent. Algebraically, the quadratic becomes a perfect square, such as \((x - 3)^2 = 0\).
If the quadratic has no real solutions (for example \((x - 2)^2 = -5\)), the graphs do not meet.
Substituting a line into a curve gives a quadratic \(ax^2 + bx + c = 0\). If it does not factorise, use
The number under the square root decides how many points there are:
| \(b^2 - 4ac\) | Solutions | Graphs |
|---|---|---|
| positive | two | the line cuts the curve twice |
| zero | one | the line is a tangent |
| negative | none | the line misses the curve |
Solving a line with a curve
- Rearrange the linear equation to make \(y\) (or \(x\)) the subject, choosing whichever avoids fractions.
- Substitute into the non-linear equation. If it is \(y = \dfrac{k}{x}\), multiply both sides by \(x\).
- Rearrange to \(ax^2 + bx + c = 0\). Divide by a common factor if there is one.
- Solve: factorise, or use the quadratic formula (exact surd answers), or complete the square.
- Pair up: find \(y\) for each \(x\) from the linear equation, then check each point in the curve.
Questions about the number of points
- Show it is a tangent: reach a perfect square, e.g. \((y - 6)^2 = 0\), and say 'exactly one point'.
- Show they do not meet: complete the square to get a square equal to a negative number, or show \(b^2 - 4ac < 0\).
- Find \(k\) for one point: write the quadratic with \(k\) in it, then make the square equal zero (or \(b^2 - 4ac = 0\)).
- One point given: if a line \(y = mx + c\) meets \(y = x^2\) where \(x = -2\), that point is \((-2, 4)\); substitute it to find \(c\), then solve for the other point.
- In context: a path \(h = -0.2x^2 + 3x\) meeting a slope \(h = x\) lands where the heights are equal; \(x = 0\) is the start, so take the other solution. A gap at a given \(x\) is the difference of the two heights.
Make \(y\) the subject of the linear equation: \(y = x + 7\). Substitute it into the parabola and move every term to one side.
| \(x + 7\) | \(=\) | \(x^2 - 2x - 3\) |
| \(0\) | \(=\) | \(x^2 - 3x - 10\) |
| \(0\) | \(=\) | \((x - 5)(x + 2)\) |
| \(x\) | \(=\) | \(5 \text{ or } x = -2\) |
Find each \(y\) from the line (the simpler equation):
| \(x = 5:\quad y\) | \(=\) | \(12\) |
| \(x = -2:\quad y\) | \(=\) | \(5\) |
Check on the parabola: \(25 - 10 - 3 = 12\ \checkmark\) and \(4 + 4 - 3 = 5\ \checkmark\)
The points are \(\boldsymbol{(5,\ 12)}\) and \(\boldsymbol{(-2,\ 5)}\).
Set the two expressions for \(y\) equal, then multiply both sides by \(x\) (\(x \neq 0\) on the hyperbola).
| \(\dfrac{6}{x}\) | \(=\) | \(2x - 4\) |
| \(6\) | \(=\) | \(2x^2 - 4x\) |
| \(0\) | \(=\) | \(2x^2 - 4x - 6\) |
| \(0\) | \(=\) | \(x^2 - 2x - 3\) |
| \(0\) | \(=\) | \((x - 3)(x + 1)\) |
| \(x = 3:\quad y\) | \(=\) | \(2(3) - 4 = 2\) |
| \(x = -1:\quad y\) | \(=\) | \(2(-1) - 4 = -6\) |
Check: \(\dfrac{6}{3} = 2\ \checkmark\) and \(\dfrac{6}{-1} = -6\ \checkmark\). Solutions \(\boldsymbol{(3,\ 2)}\) and \(\boldsymbol{(-1,\ -6)}\).
The same works for \(xy = 6\): substitute \(y = 2x - 4\) to get \(x(2x - 4) = 6\).
Substitute the line into the circle and expand \((x + 10)^2\) in full.
| \(x^2 + (x + 10)^2\) | \(=\) | \(50\) |
| \(x^2 + x^2 + 20x + 100\) | \(=\) | \(50\) |
| \(2x^2 + 20x + 50\) | \(=\) | \(0\) |
| \(x^2 + 10x + 25\) | \(=\) | \(0\) |
| \((x + 5)^2\) | \(=\) | \(0\) |
A perfect square gives only one value, \(x = -5\), so there is exactly one meeting point.
| \(y\) | \(=\) | \(-5 + 10 = 5\) |
Check: \((-5)^2 + 5^2 = 50\ \checkmark\). The line touches the circle at \(\boldsymbol{(-5,\ 5)}\) without crossing it: it is a tangent.
(i) Find the exact points of intersection of \(y = x^2 + 2\) and \(y = 2x + 4\).
(ii) For which value of \(k\) does \(y = 2x + k\) meet \(y = x^2 + 2\) at exactly one point?
(iii) For which \(k\) do they not meet?
(i) Equate and use the quadratic formula, since \(x^2 - 2x - 2\) does not factorise.
| \(x^2 + 2\) | \(=\) | \(2x + 4\) |
| \(x^2 - 2x - 2\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\dfrac{2 \pm \sqrt{4 + 8}}{2}\) |
| \(\) | \(=\) | \(\dfrac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}\) |
| \(y\) | \(=\) | \(2(1 \pm \sqrt{3}) + 4\) |
| \(\) | \(=\) | \(6 \pm 2\sqrt{3}\) |
Points: \(\boldsymbol{\left(1 + \sqrt{3},\ 6 + 2\sqrt{3}\right)}\) and \(\boldsymbol{\left(1 - \sqrt{3},\ 6 - 2\sqrt{3}\right)}\). Keep the signs paired.
(ii) Equate and complete the square:
| \(x^2 + 2\) | \(=\) | \(2x + k\) |
| \(x^2 - 2x + 2 - k\) | \(=\) | \(0\) |
| \((x - 1)^2\) | \(=\) | \(k - 1\) |
A square equals \(k - 1\). One solution needs \(k - 1 = 0\): \(\boldsymbol{k = 1}\), touching at \(x = 1\), the point \((1, 3)\).
(iii) A square cannot be negative, so there are no solutions when \(k - 1 < 0\): \(\boldsymbol{k < 1}\). (For \(k > 1\) there are two points.)
Common pitfalls
Frequently asked questions
How do you solve simultaneous equations when one is a parabola?
Make y the subject of the linear equation, substitute it into the parabola, and rearrange to a quadratic equal to zero. Solve the quadratic, then find each y value from the line.
How many solutions can a line and a parabola have?
Two, one or none. Two if the line cuts the parabola, one if it is a tangent, and none if it misses. The quadratic you get has the same number of solutions.
How do you show a line is a tangent to a curve?
Substitute the line into the curve and show the quadratic is a perfect square, such as (x minus 3) squared equals 0. There is then only one point of contact.
How do you solve a line and a circle simultaneously?
Make x or y the subject of the line, substitute into x squared plus y squared equals r squared, expand the bracket squared fully, and solve the quadratic. Find the other coordinate from the line.
What if the quadratic has no solutions?
Then the line and the curve do not meet. You can show this by completing the square and getting a square equal to a negative number.