This problem is a comprehensive test of your understanding of parabolas, tangents, normals, and circles. We will systematically use coordinate geometry formulas and crucial geometric properties to solve it.

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Problem Statement Analysis:

We are given a parabola $y^2 = 16x$ and a specific point $P(16, 16)$ on it.

1. A tangent is drawn at $P$, intersecting the parabola's axis (the x-axis) at point $A$.
2. A normal is drawn at $P$, intersecting the parabola's axis (the x-axis) at point $B$.
3. $C$ is the center of the circle that passes through points $P$, $A$, and $B$.
4. We need to find a value for $\tan\theta$, where $\theta = \angle CPB$.

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Key Concepts and Formulas:

Before diving into the solution, let's list the essential concepts and formulas we'll employ:

1. Parabola Equation: The standard form of a parabola opening to the right is $y^2 = 4ax$. Its axis of symmetry is the x-axis ($y=0$).
2. Equation of Tangent: For a parabola $y^2 = 4ax$, the equation of the tangent at a point $P(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.
3. Equation of Normal: The normal to a curve at a point is perpendicular to the tangent at that point. If the slope of the tangent is $m_T$, the slope of the normal $m_N = -1/m_T$. The equation of the normal can be found using the point-slope form: $y - y_1 = m_N(x - x_1)$.
4. Geometric Property of Tangent and Normal: For any point $P$ on a parabola, the tangent and normal drawn at $P$ intersect the axis of the parabola at points $A$ and $B$ respectively, such that $\angle APB = 90^\circ$. This is a crucial property that simplifies finding the circle's center.
5. Circle through P, A, B: Since $\angle APB = 90^\circ$, the segment $AB$ acts as the diameter of the circle passing through $P$, $A$, and $B$. Consequently, the center $C$ of this circle is the midpoint of $AB$.
6. Midpoint Formula: The midpoint of a segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
7. Slope of a Line: The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
8. Angle Between Two Lines: If two lines have slopes $m_1$ and $m_2$, the tangent of the acute angle $\alpha$ between them is given by $\tan\alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.

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Step-by-Step Solution:

1. Identify Parabola Parameters and Verify Point P

• Key Concept: The standard form of a parabola opening to the right is $y^2 = 4ax$.
• Why this step: To extract the parameter 'a' which is essential for tangent and normal equations, and to confirm that the given point $P$ lies on the parabola.
• Working: The given equation of the parabola is $y^2 = 16x$. Comparing this with the standard form $y^2 = 4ax$, we find $4a = 16$, which implies $a = 4$. The axis of the parabola $y^2 = 16x$ is the x-axis, defined by the line $y=0$. The given point is $P(16, 16)$. We verify its position on the parabola: Substitute $x=16, y=16$ into $y^2 = 16x$: $(16)^2 = 16(16)$ $256 = 256$ Since the equation holds true, point $P$ indeed lies on the parabola.
• Tip: Always verify the point lies on the curve. This simple check can prevent errors if the point was misidentified.

2. Find the Equation of the Tangent at P and Determine Point A

• Key Concept: The equation of the tangent to the parabola $y^2 = 4ax$ at a point $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.
• Why this step: We need to find the line PA and then determine where it intersects the parabola's axis ($y=0$) to locate point A.
• Working: Substitute $x_1 = 16$, $y_1 = 16$, and $a = 4$ into the tangent formula: $y(16) = 2(4)(x + 16)$ $16y = 8(x + 16)$ Divide both sides by 8 to simplify: $2y = x + 16 \quad \text{(Equation of Tangent PA)}$ Point $A$ is the intersection of the tangent with the x-axis ($y=0$). Set $y=0$ in the tangent equation: $2(0) = x + 16$ $0 = x + 16$ $x = -16$
• Result: The coordinates of point A are $(-16, 0)$.

3. Find the Equation of the Normal at P and Determine Point B

• Key Concept: The normal to a curve at a point is perpendicular to the tangent at that point. If the slope of the tangent is $m_T$, the slope of the normal $m_N = -1/m_T$. The equation of a line with slope $m$ passing through $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$.
• Why this step: We need to find the line PB and then determine where it intersects the parabola's axis ($y=0$) to locate point B.
• Working: From the tangent equation $2y = x + 16$, we can find its slope by rearranging it into the slope-intercept form $y = mx + c$: $y = \frac{1}{2}x + 8$ The slope of the tangent PA is $m_T = \frac{1}{2}$. The slope of the normal PB ($m_N$) is the negative reciprocal of $m_T$: $m_N = -\frac{1}{m_T} = -\frac{1}{1/2} = -2$ Now, use the point-slope form for the normal, with $P(16, 16)$ and $m_N = -2$: $y - 16 = -2(x - 16)$ $y - 16 = -2x + 32$ $y = -2x + 48 \quad \text{(Equation of Normal PB)}$ Point $B$ is the intersection of the normal with the x-axis ($y=0$). Set $y=0$ in the normal equation: $0 = -2x + 48$ $2x = 48$ $x = 24$
• Result: The coordinates of point B are $(24, 0)$.
• Tip: Be careful with arithmetic, especially when dealing with fractions and signs for negative reciprocals.

4. Determine the Centre C of the Circle PAB

• Key Concept: A crucial geometric property for parabolas states that the tangent and normal at any point $P$ on a parabola intersect the axis of the parabola at points $A$ and $B$ respectively, such that $\angle APB = 90^\circ$. If three points $P, A, B$ form a right angle at $P$, then the segment connecting the other two points ($AB$) must be the diameter of the circle passing through $P, A, B$.
• Why this step: This property significantly simplifies finding the center $C$. Instead of solving general circle equations, we can simply find the midpoint of the diameter $AB$.
• Working: Using the coordinates of $A(-16, 0)$ and $B(24, 0)$, we find the midpoint $C$: $C = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right)$ $C = \left(\frac{-16 + 24}{2}, \frac{0 + 0}{2}\right)$ $C = \left(\frac{8}{2}, \frac{0}{2}\right)$ $C = (4, 0)$
• Result: The coordinates of the center C are $(4, 0)$.
• Tip: Recognizing this geometric property ($\angle