The problem asks us to find the value of $k$, given that the line $y = 2x + 1$ meets the curve with the equation $kx^2 + 2y + (k - 2) = 0$ at exactly one point, and that $k$ is a positive constant.

Step-by-step solution:

1. Substitute the line equation into the curve equation: Since the line intersects the curve, we substitute $y = 2x + 1$ into the curve equation $kx^2 + 2y + (k - 2) = 0$.

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

2. Expand the terms: $kx^2 + 4x + 2 + k - 2 = 0$ Simplify the constants: $kx^2 + 4x + k = 0$

3. Solve for the condition of exactly one point: The condition that the line meets the curve at exactly one point means that the quadratic equation must have exactly one solution. This happens when the discriminant is zero.

   The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by: $\Delta = b^2 - 4ac$ For the equation $kx^2 + 4x + k = 0$, we have:

   • $a = k$
   • $b = 4$
   • $c = k$

   The discriminant is: $\Delta = 4^2 - 4(k)(k)$ Simplifying: $\Delta = 16 - 4k^2$

4. Set the discriminant equal to zero: To ensure exactly one solution, we set the discriminant equal to zero: $16 - 4k^2 = 0$ Solve for $k$: $4k^2 = 16$ $k^2 = 4$ $k = 2$ Since $k$ is given as a positive constant, the value of $k$ is $2$.

Thus, the value of $k$ is $\boxed{2}$.

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. What would happen if the discriminant were positive or negative?
2. How do we determine if two curves intersect at more than one point?
3. Can this method be applied to any curve and line intersection?
4. How would the solution change if the curve had a different general form, like $kx^2 + 3y + (k - 5) = 0$?
5. What if the line equation was $y = mx + b$ instead of $y = 2x + 1$?

Tip: The discriminant is key in determining how many solutions a quadratic equation has, making it essential in problems involving intersections between curves and lines.