Here is the elaborated, clear, and educational solution for the given problem.

Problem Statement: Let P(3, 3) be a point on the hyperbola, ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$. If the normal to it at P intersects the x-axis at (9, 0) and e is its eccentricity, then the ordered pair $(a^2, e^2)$ is equal to:

Options: (A) $\left( {{9 \over 2},2} \right)$ (B) $\left( {{3 \over 2},2} \right)$ (C) (9,3) (D) $\left( {{9 \over 2},3} \right)$

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Key Concepts and Formulas:

Before we embark on the solution, let's review the essential definitions and formulas related to hyperbolas that will be instrumental in solving this problem. A strong understanding of these fundamentals is crucial for success in coordinate geometry problems.

1. Standard Equation of a Hyperbola: For a hyperbola centered at the origin, with its transverse axis along the x-axis, the standard equation is: ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$ Here, $a$ represents the length of the semi-transverse axis (half the distance between the vertices), and $b$ represents the length of the semi-conjugate axis. Both $a^2$ and $b^2$ must be positive constants.

2. Equation of the Normal to a Hyperbola: The equation of the normal line to the hyperbola ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$ at a point $(x_1, y_1)$ lying on the hyperbola is given by: ${{{a^2}x} \over {{x_1}}} + {{{b^2}y} \over {{y_1}}} = {a^2} + {b^2}$ This is a standard formula. For problems involving eccentricity, it's often more convenient to express the right-hand side in terms of $a^2$ and $e^2$. We know that for a hyperbola, the relationship between $a$, $b$, and $e$ is $b^2 = a^2(e^2 - 1)$. Substituting this into the right-hand side of the normal equation: ${a^2} + {b^2} = {a^2} + a^2(e^2 - 1) = {a^2} + {a^2}{e^2} - {a^2} = {a^2}{e^2}$ Therefore, the equation of the normal can also be written in a very useful form directly involving eccentricity: ${{{a^2}x} \over {{x_1}}} + {{{b^2}y} \over {{y_1}}} = {a^2}{e^2}$

3. Eccentricity of a Hyperbola: The eccentricity $e$ is a key characteristic of a hyperbola, determining its shape. It is always greater than 1 for a hyperbola. The relationship between $e$, $a$, and $b$ is given by: ${e^2} = 1 + {{{b^2}} \over {{a^2}}}$ This formula allows us to connect the dimensions of the hyperbola ($a$ and $b$) to its eccentricity.

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Step-by-Step Solution:

We are given a hyperbola, a specific point P(3, 3) on it, and information about where the normal at P intersects the x-axis. Our objective is to determine the ordered pair $(a^2, e^2)$.

Step 1: Utilize the fact that P(3, 3) lies on the hyperbola.

• Why this step? If a point lies on a curve, its coordinates must satisfy the equation of that curve. This fundamental principle allows us to establish the first algebraic relationship between our unknown parameters $a^2$ and $b^2$. This is always the first step when a point on the conic section is given.
• Action: Substitute the coordinates of point $P(x_1, y_1) = (3, 3)$ into the standard equation of the hyperbola: ${{{3^2}} \over {{a^2}}} - {{{3^2}} \over {{b^2}}} = 1$ Simplifying the squares, we get: ${{9} \over {{a^2}}} - {{9} \over {{b^2}}} = 1 \quad \text{...(1)}$ This equation is crucial as it connects $a^2$ and $b^2$.

Step 2: Formulate the Equation of the Normal to the hyperbola at P(3, 3).

• Why this step? The problem provides specific information about the normal line (its intersection with the x-axis). To use this information, we first need the algebraic equation that represents this normal line. We will use the eccentricity form of the normal equation because we eventually need to find $e^2$.
• Action: Using the formula for the normal at $(x_1, y_1) = (3, 3)$: ${{{a^2}x} \over {x_1}} + {{{b^2}y} \over {y_1}} = {a^2}{e^2}$ Substitute $x_1=3$ and $y_1=3$: ${{{a^2}x} \over {3}} + {{{b^2}y} \over {3}} = {a^2}{e^2} \quad \text{...(2)}$ This equation now represents the normal line to the hyperbola at point P.

Step 3: Utilize the x-intercept of the normal to find $e^2$.

• Why this step? We are given that the normal line intersects the x-axis at the point (9, 0). Any point on the x-axis has a y-coordinate of 0. By substituting these specific coordinates into the normal's equation (Equation 2), we can eliminate $y$ and obtain an equation that allows us to directly solve for $e^2$.
• Action: Substitute the coordinates of the x-intercept $(x, y) = (9, 0)$ into Equation (2): ${{{a^2}(9)} \over {3}} + {{{b^2}(0)} \over {3}} = {a^2}{e^2}$ Simplify the terms: ${3{a^2}} + 0 = {a^2}{e^2}$ ${3{a^2}} = {a^2}{e^2}$ Since $a^2$ cannot be zero for a hyperbola (it represents the square of a semi-axis length), we can safely divide both sides by $a^2$: $3 = {e^2}$ We have successfully found the value of $e^2$: ${e^2} = 3$

Step 4: Connect $b^2$ and $a^2$ using the eccentricity.

• Why this step? We have found one of the required values ($e^2$). To find $a^2$, we need another relationship between $a^2$ and $b^2$ that we can combine with Equation (1). The eccentricity formula provides this crucial link. By substituting the value of $e^2$ we just found, we can express $b^2$ directly in terms of $a^2$.
• Action: Use the eccentricity formula ${e^2} = 1 + {{{b^2}} \over {{a^2}}}$ and substitute our derived value $e^2 = 3$: $3 = 1 + {{{b^2}} \over {{a^2}}}$ Subtract 1 from both sides: $2 = {{{b^2}} \over {{a^2}}}$ Multiply both sides by $a^2$ to solve for $b^2$: ${b^2} = 2{a^2} \quad \text{...(3)}$ This equation provides a direct and simple relationship between $b^2$ and $a^2$.

Step 5: Solve for $a^2$ using the initial condition.

• Why this step? We now have two independent equations relating $a^2$ and $b^2$: Equation (1) from the point P lying on the hyperbola, and Equation (3) derived from the eccentricity. We can now substitute the expression for $b^2$ from Equation (3) into Equation (1). This will result in a single equation with only $a^2$ as the unknown, allowing us to solve for its value.
• Action: Substitute $b^2 = 2a^2$ from Equation (3) into Equation (1): ${{9} \over {{a^2}}} - {{9} \over {2{a^2}}} = 1$ To combine the terms on the left side, find a common denominator, which is $2a^2$: ${{2 \times 9} \over {2{a^2}}} - {{9} \over {2{a^2}}} = 1$ ${{18} \over {2{a^2}}} - {{9} \over {2{a^2}}} = 1$ Combine the numerators: ${{18 - 9} \over {2{a^2}}} = 1$ ${{9} \over {2{a^2}}} = 1$ Multiply both sides by $2a^2$: $9 = 2{a^2}$ Divide by 2 to solve for $a^2$: ${a^2} = {9 \over 2}$

Step 6: Form the ordered pair $(a^2, e^2)$ and Select the Option.

• Why this step? The question explicitly asks for the ordered pair $(a^2, e^2)$. We have successfully calculated both values in the preceding steps. This is the final step to present our answer in the required format and match it with the given options.
• Action: Combine the results from Step 3 ($e^2 = 3$) and Step 5 ($a^2 = 9/2$): $$$$