Problem 187

1. Since curve C is the set of points for which the absolute difference of the distances to two fixed points F₁(-2, 0) and F₂(2, 0) is equal to a constant length 2$\sqrt {3}$, by the definition of a hyperbola, curve C is a hyperbola with foci F₁(-2, 0) and F₂(2, 0) and a real axis length of 2$\sqrt {3}$. Therefore, the equation of curve C is $\frac{x^{2}}{3} - y^{2} = 1$.

2. Since line l passes through the point (0, 1), when the slope of line l does not exist, the line is x = 0, which is not valid. When the slope k of line l exists, let the equation of line l be y = kx + 1. When k = $± \frac{\sqrt{3}}{3}$, line l is y = $± \frac{\sqrt{3}}{3}$x + 1, which intersects curve C at only one focus. Solving the system of equations:

$$\begin{cases} x^{2} - 3y^{2} = 3 \\ y = kx + 1 \end{cases}$$

We get (1 - 3k²)x² - 6kx - 6 = 0. When 1 - 3k² ≠ 0, the discriminant Δ = 36k² + 24(1 - 3k²) = 0, and we find that k = ±2. Therefore, line l has only one common point with curve C, and the equation of line l is y = ±2x + 1.

In summary, the equation of line l is y = $± \frac{\sqrt{3}}{3}$x + 1 or y = ±2x + 1.

3. When a = 1, the equation of curve C is $x^{2} - \frac{y^{2}}{3} = 1$. Solving the system of equations:

$$\begin{cases} x^{2} - \frac{y^{2}}{3} = 1 \\ y = kx + 2 \end{cases}$$

We get (3 - k²)x² - 4kx - 4 = 0. Let A(x₁, y₁) and B(x₂, y₂) be the intersection points. Then, x₁ + x₂ = $\frac{4k}{3 - k^{2}}$ and x₁x₂ = $- \frac{4}{3 - k^{2}}$. Also, y₁y₂ = (kx₁ + 2)(kx₂ + 2) = k²x₁x₂ + k(x₁ + x₂) + 4 = $- \frac{4k^{2}}{3 - k^{2}} + \frac{4k^{2}}{3 - k^{2}} + 4 = 4$.

Since OA is perpendicular to OB, x₁x₂ + y₁y₂ = 4 - $\frac{4}{3 - k^{2}}$ = 0. Therefore, 3 - k² = 1, and we find that k = $± \sqrt{2}$.

The final answers are:

1. The equation of curve C is $\boxed{\frac{x^{2}}{3} - y^{2} = 1}$.
2. The equation of line l is $\boxed{y = ± \frac{\sqrt{3}}{3}x + 1 \text{ or } y = ±2x + 1}$.
3. The value of k is $\boxed{k = ± \sqrt{2}}$.