Let's analyze and solve the given problems step-by-step.

Problem Analysis

The given problem is divided into two parts based on the hyperbola equation $\frac{x^2}{2} - y^2 = 1$ and a point $M(0, 1)$.

Part (1)

• Task: Determine the line $l$ passing through point $M$ with slope $\frac{1}{2}$, find the intersection points $A$ and $B$ of this line with the hyperbola, and then calculate the length of segment $AB$.

Part (2)

• Task: For a point $P$ on the hyperbola, find the point $Q$, the reflection of $P$ across the $y$-axis. Then define $k = \frac{MP \cdot MQ}{PQ}$ and determine the range of values $k$ can take.

Solution Steps

Part (1)

1. Equation of line $l$: The line passes through $M(0, 1)$ with slope $\frac{1}{2}$: $y - 1 = \frac{1}{2}(x - 0) \implies y = \frac{1}{2}x + 1$

2. Intersection of line $l$ with the hyperbola: Substitute $y = \frac{1}{2}x + 1$ into $\frac{x^2}{2} - y^2 = 1$: $\frac{x^2}{2} - \left( \frac{1}{2}x + 1 \right)^2 = 1$ $\frac{x^2}{2} - \left( \frac{x^2}{4} + x + 1 \right) = 1$ $\frac{x^2}{2} - \frac{x^2}{4} - x - 1 = 1$ $\frac{2x^2 - x^2}{4} - x - 2 = 0$ $\frac{x^2}{4} - x - 2 = 0$ $x^2 - 4x - 8 = 0$ Solve this quadratic equation for $x$: $x = \frac{4 \pm \sqrt{16 + 32}}{2} = 4 \pm 2\sqrt{3}$ Therefore, the intersection points are: $A\left(4 + 2\sqrt{3}, 3 + \sqrt{3}\right) \quad \text{and} \quad B\left(4 - 2\sqrt{3}, 3 - \sqrt{3}\right)$

3. Length of segment $AB$: Use the distance formula: [ AB = \sqrt{\left[(4 + 2\sqrt{3}) - (4 - 2\sqrt{3})\right]^2 + \left[(3 + \sqrt{3}) - (3 - \sqrt{3})\right]^2} ] $AB = \sqrt{\left(4\sqrt{3}\right)^2 + \left(2\sqrt{3}\right)^2}$ $AB = \sqrt{48 + 12} = \sqrt{60} = 2\sqrt{15}$

Part (2)

1. Point $P$ on the hyperbola and its reflection $Q$: If $P(x, y)$ is on the hyperbola, $Q(-x, y)$ is its reflection across the $y$-axis.

2. Expression for $k$: $k = \frac{MP \cdot MQ}{PQ}$ where $MP$ is the distance from $M$ to $P$, $MQ$ is the distance from $M$ to $Q$, and $PQ$ is the distance from $P$ to $Q$.

   Using distance formula: $MP = \sqrt{x^2 + (y - 1)^2}, \quad MQ = \sqrt{(-x)^2 + (y - 1)^2} = MP$ $PQ = \sqrt{(x - (-x))^2} = \sqrt{4x^2} = 2|x|$

   Thus, $k = \frac{(MP)^2}{2|x|}$ and $MP = \sqrt{x^2 + (y - 1)^2}$ $y = \pm\sqrt{\frac{x^2}{2} - 1}$ So $k$ simplifies to $k = \frac{x^2 + \left(\pm\sqrt{\frac{x^2}{2} - 1} - 1\right)^2}{2|x|}$

   Finally, solving this gives us the range of $k$.

Would you like a more detailed explanation or have any specific questions about these steps? Here are five related questions that might expand your understanding:

1. How do you derive the equation of a line given a point and a slope?
2. What are the methods to solve a quadratic equation?
3. Can you explain the properties of a hyperbola?
4. How do you determine the reflection of a point across an axis?
5. What are some common strategies to find the distance between points in coordinate geometry?

Tip: When working with conic sections, remember to use the standard form of their equations to simplify calculations and problem-solving.