A Line and a Curve for Every Value of a Constant

To show that a line meets, touches or never meets a curve for every value of a constant, reduce the two equations to one quadratic equation and show that its discriminant \(D\) (also written \(\Delta\)) has the required sign for every value of the constant.

Why: every line \(y=kx+5\) passes through \((0,\,5)\), and that point lies above the curve \(y=x^2+2x+2\), which passes through \((0,\,2)\). A line through a point inside the curve must cross it twice, whatever its gradient. The discriminant proves this for every \(k\).

Method

1. Substitute the line into the curve and collect all terms on one side as \(ax^2+bx+c=0\).
2. If the constant appears in \(a\), substitute the value that makes \(a=0\) and solve what is left separately: a line and a line meet once, never if they are parallel, or everywhere if they coincide.
3. Write \(D=b^2-4ac\) as a square, a square plus a positive number, or minus a square minus a positive number.
4. Use the conditions from the Intersection of a Line and a Curve card: meets needs \(D\ge0\), two distinct points \(D>0\), never meets \(D<0\). State the sign for every value and finish with the required sentence.

Example

Show that the line \(y=kx+5\) meets the curve \(y=x^2+2x+2\) at two distinct points for all real values of \(k\).

\(\begin{aligned} x^2+2x+2&=kx+5 \\ x^2+(2-k)x-3&=0 \end{aligned}\)

\(\begin{aligned} D&=(2-k)^2-4(1)(-3) \\ &=(k-2)^2+12>0 \end{aligned}\)

Since \(D>0\) for every real \(k\), the line meets the curve at two distinct points for all real values of \(k\).

Exam check \(D=(k-2)^2\) shows that a line meets a curve, not that it meets it at two points: at \(k=2\) the line is a tangent. To show a line never meets a curve, \(D\) must be negative for every value, for example \(D=-(k^2+3)\), and any zero case must leave an equation with no solution.