This problem requires a deep understanding of hyperbolas, including their standard equation, eccentricity, and the equations of tangents and normals at a given point. Our strategy will involve systematically determining the hyperbola's parameters, finding the equations of the tangent and normal at the specified point P, locating their intersections with the conjugate axis (Q and R), and finally calculating the distance between these intersection points.

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1. Determine the Hyperbola's Parameters ($a^2$ and $b^2$)

To work with the hyperbola, we first need to establish its specific equation by finding the values of $a^2$ and $b^2$. We will use the two pieces of information provided: the point P lies on the hyperbola, and its eccentricity.

Key Concepts:

• The standard equation of a hyperbola with its transverse axis along the x-axis is given by: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ where $a$ is the semi-transverse axis length and $b$ is the semi-conjugate axis length.
• The eccentricity $e$ of a hyperbola is related to $a$ and $b$ by the formula: $b^2 = a^2(e^2 - 1)$

Step 1.1: Use the given point $P\left( { - 2\sqrt 6 ,\sqrt 3 } \right)$ to form an equation. Why this step? Since the point P lies on the hyperbola, its coordinates must satisfy the hyperbola's equation. This will give us a relationship between $a^2$ and $b^2$.

Substitute $x = -2\sqrt{6}$ and $y = \sqrt{3}$ into the hyperbola equation: $\frac{(-2\sqrt{6})^2}{a^2} - \frac{(\sqrt{3})^2}{b^2} = 1$ $\frac{4 \times 6}{a^2} - \frac{3}{b^2} = 1$ $\frac{24}{a^2} - \frac{3}{b^2} = 1 \quad \text{...(1)}$

Step 1.2: Use the given eccentricity $e = \frac{\sqrt{5}}{2}$ to form another equation. Why this step? The eccentricity formula provides an independent relationship between $a^2$ and $b^2$. With two such relationships, we can solve for the unique values of $a^2$ and $b^2$.

Substitute $e = \frac{\sqrt{5}}{2}$ into the eccentricity relation $b^2 = a^2(e^2 - 1)$: $b^2 = a^2\left( \left(\frac{\sqrt{5}}{2}\right)^2 - 1 \right)$ $b^2 = a^2\left( \frac{5}{4} - 1 \right)$ $b^2 = a^2\left( \frac{5-4}{4} \right)$ $b^2 = \frac{a^2}{4}$ This implies $a^2 = 4b^2$.

Step 1.3: Solve the system of equations for $a^2$ and $b^2$. Why this step? To find the specific numerical values for $a^2$ and $b^2$ that define our particular hyperbola.

We have two equations:

1. $\frac{24}{a^2} - \frac{3}{b^2} = 1$
2. $a^2 = 4b^2$

Substitute $a^2 = 4b^2$ from the second equation into equation (1): $\frac{24}{4b^2} - \frac{3}{b^2} = 1$ $\frac{6}{b^2} - \frac{3}{b^2} = 1$ $\frac{3}{b^2} = 1$ $b^2 = 3$ Now, substitute the value of $b^2$ back into $a^2 = 4b^2$: $a^2 = 4 \times 3$ $a^2 = 12$ Thus, the specific equation of the hyperbola is $\frac{x^2}{12} - \frac{y^2}{3} = 1$.

Tip: It's always a good practice to double-check your calculated values of $a^2$ and $b^2$ by substituting them back into the original conditions. For $P(-2\sqrt{6}, \sqrt{3})$: $\frac{(-2\sqrt{6})^2}{12} - \frac{(\sqrt{3})^2}{3} = \frac{24}{12} - \frac{3}{3} = 2 - 1 = 1$. This confirms our values are correct.

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2. Find the Intersection Point Q (Tangent with Conjugate Axis)

Key Concepts:

• The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at a point $P(x_1, y_1)$ on it is given by: $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$
• For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (transverse axis along x-axis), its conjugate axis is the y-axis, which has the equation $x=0$.

Step 2.1: Write down the equation of the tangent at P. Why this step? We need the explicit equation of the tangent line to find where it intersects the conjugate axis.

The point of tangency is $P(x_1, y_1) = (-2\sqrt{6}, \sqrt{3})$, and we have $a^2=12$, $b^2=3$. Substitute these values into the tangent equation formula: $\frac{x(-2\sqrt{6})}{12} - \frac{y(\sqrt{3})}{3} = 1$ $\frac{-x\sqrt{6}}{6} - \frac{y\sqrt{3}}{3} = 1$ To clear the denominators, multiply the entire equation by 6: $-x\sqrt{6} - 2y\sqrt{3} = 6 \quad \text{...(2)}$

Step 2.2: Find the intersection with the conjugate axis to determine Q. Why this step? The problem defines Q as the intersection of the tangent with the conjugate axis ($x=0$). We substitute $x=0$ into the tangent equation to find its y-coordinate.

Substitute $x=0$ into the tangent equation (2): $-(0)\sqrt{6} - 2y\sqrt{3} = 6$ $-2y\sqrt{3} = 6$ $y = \frac{6}{-2\sqrt{3}} = \frac{3}{-\sqrt{3}}$ To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$: $y = -\frac{3\sqrt{3}}{3} = -\sqrt{3}$ So, the point Q is $(0, -\sqrt{3})$.

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3. Find the Intersection Point R (Normal with Conjugate Axis)

Key Concepts:

• The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at a point $P(x_1, y_1)$ on it is given by: $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$
• The conjugate axis is the y-axis, with equation $x=0$.

Step 3.1: Write down the equation of the normal at P. Why this step? Similar to the tangent, we need the explicit equation of the normal line to find where it intersects the conjugate axis.

Using the point $P(x_1, y_1) = (-2\sqrt{6}, \sqrt{3})$, and $a^2=12$, $b^2=3$:

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