This solution will guide you step-by-step through the problem, elaborating on each concept and calculation.

1. Understanding the Problem and Identifying Key Concepts

The problem asks us to find the sum of the product of all possible 'a' values and the product of all possible 'b' values for a point $P(a, b)$ on the parabola $y^2 = 8x$. The crucial condition is that the tangent to the parabola at $P$ passes through the center of a given circle.

To solve this, we will need the following key concepts and formulas:

• Centre of a Circle: For a circle given by the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$, its center is at the point $(-g, -f)$.
• Equation of Tangent to a Parabola: For a parabola of the form $y^2 = 4kx$, the equation of the tangent at a point $P(x_1, y_1)$ lying on the parabola is given by $yy_1 = 2k(x+x_1)$.
• Point on Parabola Condition: If a point $P(x_1, y_1)$ lies on the parabola $y^2 = 4kx$, it must satisfy the parabola's equation, i.e., $y_1^2 = 4kx_1$.
• Vieta's Formulas: For a quadratic equation $px^2 + qx + r = 0$, if $x_1$ and $x_2$ are its roots, then the sum of roots $x_1+x_2 = -q/p$ and the product of roots $x_1x_2 = r/p$. These formulas are extremely useful for finding products or sums of roots without explicitly calculating the roots themselves.

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2. Step-by-Step Solution

Step 1: Determine the Centre of the Given Circle

• Concept Used: Centre of a Circle.
• Explanation: The tangent line at point $P$ passes through the center of the given circle. Therefore, our first step is to find the coordinates of this center.
• Working: The equation of the circle is $x^2 + y^2 - 10x - 14y + 65 = 0$. We compare this to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$. By comparing coefficients, we have: $2g = -10 \implies g = -5$ $2f = -14 \implies f = -7$ The center of the circle is $(-g, -f)$. So, the center $C = (-(-5), -(-7)) = (5, 7)$.

Step 2: Find the Equation of the Tangent to the Parabola at Point P(a, b)

• Concept Used: Equation of Tangent to a Parabola.
• Explanation: We need the equation of the tangent line because we are given a condition about it (passing through the circle's center). The general formula for the tangent is applied directly.
• Working: The equation of the parabola is $y^2 = 8x$. We compare this to the standard form $y^2 = 4kx$. Here, $4k = 8$, which implies $k = 2$. The point $P$ on the parabola is given as $(a, b)$. Using the tangent formula $yy_1 = 2k(x+x_1)$, and substituting $(x_1, y_1) = (a, b)$ and $k=2$: $yb = 2(2)(x+a)$ $yb = 4(x+a) \quad \text{(Equation 1)}$

Step 3: Apply the Condition that the Tangent Passes Through the Circle's Centre

• Concept Used: A point lies on a line if its coordinates satisfy the line's equation.
• Explanation: Since the tangent line (Equation 1) passes through the center of the circle $(5, 7)$, the coordinates of the center must satisfy the tangent's equation. This will give us a relationship between $a$ and $b$.
• Working: Substitute $x=5$ and $y=7$ (the coordinates of the center) into Equation 1: $7b = 4(5+a)$ $7b = 20 + 4a \quad \text{(Equation 2)}$

Step 4: Use the Condition that Point P(a, b) Lies on the Parabola

• Concept Used: Point on Parabola Condition.
• Explanation: The point $P(a, b)$ is on the parabola $y^2 = 8x$. This provides another essential relationship between $a$ and $b$, which we can use to form a system of equations.
• Working: Since $P(a, b)$ lies on $y^2 = 8x$, its coordinates must satisfy the parabola's equation: $b^2 = 8a \quad \text{(Equation 3)}$

Step 5: Solve the System of Equations for Possible Values of 'a' and 'b'

• Concept Used: Substitution method for solving systems of equations.

• Explanation: We now have two equations (Equation 2 and Equation 3) with two variables ($a$ and $b$). We will solve this system to find the possible values for $a$ and $b$. It's often easier to express one variable in terms of the other and substitute.

• Working: From Equation 3, we can express $a$ in terms of $b$: $a = \frac{b^2}{8}$ Substitute this expression for $a$ into Equation 2: $7b = 20 + 4 \left(\frac{b^2}{8}\right)$ $7b = 20 + \frac{b^2}{2}$ To eliminate the fraction, multiply the entire equation by 2: $14b = 40 + b^2$ Rearrange the terms to form a standard quadratic equation in $b$: $b^2 - 14b + 40 = 0$

  This quadratic equation will give us the possible values for $b$. Let these values be $b_1$ and $b_2$. We can solve this quadratic by factoring or using the quadratic formula: $(b-4)(b-10) = 0$ So, the possible values for $b$ are $b_1 = 4$ and $b_2 = 10$.

  Now, we find the corresponding values for $a$ using $a = b^2/8$: For $b_1 = 4$: $a_1 = \frac{4^2}{8} = \frac{16}{8} = 2$ For $b_2 = 10$: $a_2 = \frac{10^2}{8} = \frac{100}{8} = \frac{25}{2}$

  Thus, the possible points $P(a, b)$ are $(2, 4)$ and $(25/2, 10)$.

Step 6: Calculate A (Product of all possible values of 'a') and B (Product of all possible values of 'b')

• Concept Used: Vieta's Formulas.
• Explanation: Instead of explicitly finding the roots and then multiplying, Vieta's formulas provide a direct way to find the product of roots from the coefficients of the quadratic equation.
• Working:
  • For B (product of possible values of $b$): The quadratic equation for $b$ was $b^2 - 14b + 40 = 0$. Let the possible values of $b$ be $b_1$ and $b_2$. Using Vieta's formulas, the product of roots $b_1 b_2 = \frac{\text{constant term}}{\text{coefficient of } b^2} = \frac{40}{1} = 40$. Therefore, $B = 40$. *(Alternatively, using the values we found: $