The diagram shows the graphs of $y = 2x - 1$ and $y = kx^2$, where $k$ is a positive constant.
The graphs intersects at two distinct points $A$ and $B$.

1. Form a quadratic equation by equating the two functions

At the points of intersection,
$kx^2 = 2x - 1$
$kx^2 - 2x + 1 = 0$

2. Apply the discriminant condition for two distinct real roots

Since the graphs intersect at two distinct points, the quadratic equation has two distinct real roots. Therefore, the discriminant must be greater than zero ($b^2 - 4ac > 0$).
$(-2)^2 - 4(k)(1) > 0$
$4 - 4k > 0$

3. Solve the inequality to show the required result

$4 > 4k$
$k < 1$ (shown)

1. Substitute $k = 1$ and find the intersection of the two graphs.

For $k = 1$, the curve is $y = x^2$.
Equating the two equations:
$x^2 = 2x - 1$
$x^2 - 2x + 1 = 0$

2. Determine the nature of the roots and state the relationship.

Factorising the quadratic equation:
$(x - 1)^2 = 0$
Since there is only one repeated root ($x = 1$), the line intersects the curve at exactly one point.
Therefore, the line $y = 2x - 1$ is a tangent to the curve $y = x^2$.