This problem requires a solid understanding of the standard form of a hyperbola, its key parameters like semi-transverse axis, semi-conjugate axis, and eccentricity, and the equations of its directrices. We will use these concepts to first determine the specific equation of the hyperbola and then verify which of the given points lies on it.

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Key Concepts: Hyperbola Standard Form, Eccentricity, and Directrices

Let's begin by recalling the fundamental properties of a hyperbola whose transverse axis lies along the x-axis. This orientation is suggested by the given equation $k x^{2}-y^{2}=6$, where the $x^2$ term has a positive coefficient (assuming $k>0$) and the $y^2$ term has a negative coefficient.

1. Standard Form: The equation of such a hyperbola is given by: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

   • Here, $a$ represents the length of the semi-transverse axis, and $b$ represents the length of the semi-conjugate axis.
   • The vertices of this hyperbola are located at $(\pm a, 0)$.
   • The transverse axis is along the x-axis, and the conjugate axis is along the y-axis.

2. Eccentricity ($e$): Eccentricity is a crucial parameter for any conic section that defines its shape. For a hyperbola, $e > 1$. It is related to $a$ and $b$ by the formula: $e = \sqrt{1 + \frac{b^2}{a^2}} \quad \text{or equivalently} \quad e^2 = 1 + \frac{b^2}{a^2}$ This formula can also be written as $b^2 = a^2(e^2 - 1)$.

3. Directrices: A hyperbola has two directrices, which are lines perpendicular to the transverse axis. For a hyperbola in the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the equations of the directrices are: $x = \pm \frac{a}{e}$ These lines are fundamental to the definition of a hyperbola: for any point on the hyperbola, the ratio of its distance from a focus to its distance from the corresponding directrix is equal to the eccentricity $e$.

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Step-by-Step Solution

Step 1: Convert the Given Hyperbola Equation to Standard Form

The given equation of the hyperbola is $k x^{2}-y^{2}=6$. Our first objective is to transform this equation into the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. This allows us to directly identify the values of $a^2$ and $b^2$.

To achieve this standard form, we divide the entire equation by the constant on the right-hand side, which is 6: $\frac{k x^{2}}{6} - \frac{y^{2}}{6} = \frac{6}{6}$ $\frac{x^{2}}{6/k} - \frac{y^{2}}{6} = 1$

Now, by comparing this to the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we can identify the values of $a^2$ and $b^2$: $a^2 = \frac{6}{k} \quad \text{and} \quad b^2 = 6$

Important Note: For a hyperbola with its transverse axis along the x-axis (as implied by the standard form we chose), $a^2$ must be a positive real number. Therefore, $a^2 = \frac{6}{k} > 0$, which implies that $k$ must be a positive value ($k > 0$). This condition will be crucial when solving for $k$.

Step 2: Utilize the Directrix Information

The problem states that the line $x-1=0$, which simplifies to $x=1$, is a directrix of the hyperbola. From our key concepts, we know that the directrices of a hyperbola in the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $x = \pm \frac{a}{e}$.

Since $x=1$ is a directrix, we must have: $\frac{a}{e} = 1$ This implies a direct and crucial relationship between $a$ and $e$: $a = e$ This equation will be instrumental in solving for the unknown constant $k$.

Step 3: Express Eccentricity in Terms of $a^2$ and $b^2$

We know the fundamental formula for the eccentricity $e$ of a hyperbola with its transverse axis along the x-axis: $e^2 = 1 + \frac{b^2}{a^2}$ Now, we substitute the expressions for $a^2$ and $b^2$ that we found in Step 1 ($a^2 = \frac{6}{k}$ and $b^2 = 6$) into this eccentricity formula: $e^2 = 1 + \frac{6}{6/k}$ $e^2 = 1 + \frac{6k}{6}$ $e^2 = 1 + k$

Step 4: Solve for the Constant 'k'

We now have two important relationships involving $a$, $e$, and $k$: (i) $a = e$ (from the directrix information, Step 2) (ii) $e^2 = 1+k$ (from the eccentricity formula, Step 3) (iii) $a^2 = \frac{6}{k}$ (from the standard form conversion, Step 1)

From relation (i), squaring both sides gives us $a^2 = e^2$. Now, we can substitute the expressions for $a^2$ from (iii) and $e^2$ from (ii) into this equation: $\frac{6}{k} = 1+k$

Next, we solve this algebraic equation for $k$: $6 = k(1+k)$ $6 = k + k^2$ Rearranging this into a standard quadratic equation form: $k^2 + k - 6 = 0$ We can factor this quadratic equation: $(k+3)(k-2) = 0$ This gives two possible values for $k$: $k = -3$ or $k = 2$.

Common Mistake Alert: Recall the condition we established in Step 1: for $a^2 = \frac{6}{k}$ to be positive (which is necessary for $a$ to represent a real length), $k$ must be positive. Therefore, $k=-3$