Tangent and normal are drawn at P(16,16) on the parabola y²=16x, which intersect the axis of the parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and ∠CPB = θ, then a value of tanθ is

PAB is a right-angled triangle with the right angle at P. So, the circle circumscribing right-angled triangle APB will have hypotenuse (AB) as diameter as we know.
Since C is the center of the circle, so C will be the midpoint of AB.
Now given curve is y²=16x ⇒ 2y(dy/dx) = 16 ⇒ dy/dx = 8/y
|(dy/dx)|_(16,16) = 8/16 = 1/2
Equation of PA: (y-16) = 1/2(x-16) ⇒ 2y - 32 = x - 16 ⇒ -x + 2y = 16
Equation of PB: (y-16) = -2(x-16) ⇒ y - 16 = -2x + 32 ⇒ 2x + y = 48
So, A = (-16, 0), B = (24, 0)
C = ((-16+24)/2, 0) = (4, 0)
So, slope of CP is 16/-4 = -4
We know, the slope of PB = -2
m1 = -4, m2 = -2
tanθ = |(m1-m2)/(1+m1m2)| = |(-4+2)/(1+(-4)(-2))| = |-2/9|
Slope of CP is 16/(-4) = -4
Slope of PB = -2
tanθ = |(m1 - m2)/(1 + m1m2)| = |(-4 - (-2))/(1 + (-4)(-2))| = |-2/9|
Let m1 be the slope of CP and m2 be the slope of PB.
m1 = -4
m2 = -2
tan θ = |(m1 - m2) / (1 + m1m2)| = |(-4 - (-2)) / (1 + (-4)(-2))| = |-2/9|
The slope of CP is (16-0)/(-4-0) = -4
The slope of PB is (0-16)/(24-16) = -2
Let m1 = -4 and m2 = -2
tan θ = |(m1 - m2)/(1 + m1m2)| = |(-4 - (-2))/(1 + (-4)(-2))| = |-2/9|
However, this calculation is incorrect. Let's reconsider the slopes.
Slope of CP = (16-0)/(4-0) = 4
Slope of PB = -2
tan θ = |(4 - (-2))/(1 + 4(-2))| = 6/-7 = -6/7
Slope of CP = 16-0 / 4-0 = 4
Slope of PB = -2
tan θ = |(4-(-2))/(1+4(-2))| = 6/-7 = -6/7 (Incorrect)
Slope of CP = 16/-4 = -4
Slope of PB = -2
tan θ = |(-4 - (-2))/(1 + (-4)(-2))| = |-2/9|
This calculation is also incorrect. Let's use the coordinates directly.
CP has slope (16-0)/(16-4) = 16/12 = 4/3
PB has slope (16-0)/(16-24) = -2
tan θ = |(4/3 - (-2))/(1 + (4/3)(-2))| = |(10/3)/(-5/3)| = |-2| = 2