Review question

# When does $y=kx$ intersect the parabola $y=(x-1)^2$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9614

## Solution

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. $-4\le k\le0$.

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 $k \le -4$.

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).