When does $y=kx$ intersect the parabola $y=(x-1)^2$?

When does $y=kx$ intersect the parabola $y=(x-1)^2$?

Question

The values of $k$ for which the line $y=kx$ intersects the parabola $y=(x-1)^2$ are precisely

1. $k\le0$,
2. $k\ge-4$,
3. $k\ge0 \hspace{2mm} \textrm{or} \hspace{2mm} k\le-4$,
4. $-4\le k\le0$.

Suggestion

The values of $k$ for which the line $y=kx$ intersects the parabola $y=(x-1)^2$ are precisely…

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/Tu6DZjV2/width/600/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

Solution

The line $y=kx$ intersects the parabola $y=(x-1)^2$ when the equation

$$(x-1)^2 = kx$$

has real solutions.

Rearranging this equation gives

$$x^2 - (k+2)x + 1 =0,$$

which has discriminant $(k+2)^2-4$.

For the quadratic to have real solutions we need the discriminant to be non-negative, that is $(k+2)^2 \geq 4$, so we need either

$$\begin{align*} k+2 &\ge 2 \\ \textrm{or}\quad k+2 &\le -2, \end{align*}$$

i.e., $k \ge 0$ or $k \le -4$.

So the answer is (c).

Alternatively a sketch shows us what is happening.

It’s clear from the sketch that any line with positive gradient cuts the parabola, and that $y = -0.01x$ certainly does not. So the only possible option is (c).