When a Line Meets, Touches or Misses a Curve

Where a line meets a curve

The line y = 2x and the curve $y=x^2$y = x² meet at two points. At a meeting point the two y values agree, so $x^2=2x$x² = 2x there. Gather on one side and factor: $x^2-2x=0$x² − 2x = 0, so x(x − 2) = 0, and x = 0 or x = 2. The meeting points are (0, 0) and (2, 4).

Each meeting point comes from a root of that quadratic equation. So the number of real roots is the number of meeting points.

The curve is $y=x^2$y = x², and the straight line is y = 2x. The line cuts the curve at (0, 0) and at (2, 4).

Slide the line up and down

Now let the line be y = 2x + k, where k is a number that moves the line up or down without changing how steep it is. At a meeting point, $x^2=2x+k$x² = 2x + k. Gathering on one side gives $x^2-2x-k=0$x² − 2x − k = 0.

You do not need to solve this to count its roots. The discriminant, $b^2-4ac$b² − 4ac, does that. Here a = 1, b = −2 and c = −k, so $b^2-4ac={\left(-2\right)}^2-4\times 1\times \left(-k\right)=4+4k$b² − 4ac = (−2)² − 4 × 1 × (−k) = 4 + 4k.

Cuts, touches or misses

When k = 3, the discriminant is 4 + 12 = 16, which is above 0. The equation $x^2-2x-3=0$x² − 2x − 3 = 0 has two real roots, x = 3 and x = −1, so the line cuts the curve twice.

When k = −1, the discriminant is 4 − 4 = 0. The equation is $x^2-2x+1=0$x² − 2x + 1 = 0, which is ${\left(x-1\right)}^2=0$(x − 1)² = 0, so it has one repeated root, x = 1. The two meeting points have slid together into one, at (1, 1). The line touches the curve there without crossing it, and a line that touches a curve like this is a tangent to it.

When k = −3, the discriminant is 4 − 12 = −8, which is below 0. The equation has no real roots, so the line misses the curve altogether.

The line is y = 2x + k, and the handle moves k. The discriminant 4 + 4k is worked out as you drag. Slide the line down until it only just touches the parabola: that happens at k = −1, where the discriminant is 0.

At k = −3 the straight line, y = 2x − 3, passes below the curve $y=x^2$y = x² and never meets it.

One test, three answers

Substitute the line into the curve, gather the terms on one side, and work out the discriminant of the quadratic equation that results. If it is above 0, the line cuts the curve at two points. If it is exactly 0, the line touches the curve at one point and is a tangent. If it is below 0, the line misses the curve.

It does not matter which side you gather the terms on. Multiplying a quadratic equation by −1 changes the signs of a, b and c, but $b^2-4ac$b² − 4ac stays the same.

Finding k

The test can be run backwards. For y = 2x + k to be a tangent to $y=x^2$y = x², the discriminant must be 0: 4 + 4k = 0, so k = −1.

For the line to cut the curve twice, the discriminant must be above 0: 4 + 4k > 0, so k > −1. For it to miss, 4 + 4k < 0, so k < −1.

Another curve

Does the line y = 3x − 5 meet the curve $y=x^2+x$y = x² + x? Set the y values equal: $x^2+x=3x-5$x² + x = 3x − 5. Gather on one side: $x^2-2x+5=0$x² − 2x + 5 = 0. Here a = 1, b = −2 and c = 5, so the discriminant is 4 − 20 = −16. It is below 0, so the line misses the curve.

The usual mistakes

Working out the discriminant of the curve on its own. The discriminant of $x^2$x² is 0, but that says only that $y=x^2$y = x² touches the x-axis. The line comes into it through the substitution, so use the equation $x^2-2x-k=0$x² − 2x − k = 0.

Getting the sign of c wrong. In $x^2-2x-k=0$x² − 2x − k = 0 the constant term is −k, so −4ac is −4 × 1 × (−k) = +4k.

Using a discriminant above 0 for a tangent. A tangent touches the curve at exactly one point, so its discriminant is exactly 0.

The length between two points: Pythagoras' theorem

Part (b) of the next application asks for the length of a straight beam between two points. Going from one point to the other, count how far across and how far up it goes. Those two distances are the two shorter sides of a right-angled triangle, and the beam is its longest side, the hypotenuse.

Pythagoras' theorem says that the squares of the two shorter sides add up to the square of the hypotenuse. For sides 3 and 4, the hypotenuse is $\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$√(3² + 4²) = √(9 + 16) = √25 = 5.