图中显示了 $y = 2x - 1$ 和 $y = kx^2$ 的图像，其中 $k$ 是一个正常数。这两个图像相交于两个不同的点 $A$ 和 $B$。

1. Form a quadratic equation by equating the two given equations

$kx^2 = 2x - 1$
$kx^2 - 2x + 1 = 0$

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 $b^2 - 4ac > 0$.
$(-2)^2 - 4(k)(1) > 0$
$4 - 4k > 0$

3. Solve the inequality to show the required result

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

1. Substitute $k = 1$ and equate the two equations to find the points of intersection

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

2. Solve the quadratic equation or evaluate its discriminant to determine 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 $y = x^2$.