Official solution
As equation of tangent PA at (x1, y1) on the parabola y2 = 4ax,
yy1 = 2a (x + x1)
here (x1, y1) = ( 16, 16)
y . 16 = 2.4 (x + 16)
\[ \Rightarrow \] 2y = x + 16 .....(1)
At pont A value of y = 0
putting y = 0 in equation (1) we get,
0 = x + 16
\[ \Rightarrow \] x = - 16
\[\therefore\,\,\,\] Coordinate of point A = (- 16, 0)
Slope of line P A :
As\[\,\,\,\,\] 2y = x + 16
\[\therefore\,\,\,\] y = \[{1 \over 2}\,\] x + 8
\[\therefore\,\,\,\] Slope (m) = \[{1 \over 2}\,\]
Let slope of perpendicular line PB passing through point p(16, 16) = m'
\[\therefore\,\,\,\] m m' = - 1
\[ \Rightarrow \] \[{1 \over 2}\] \[ \times \] m' = - 1
\[ \Rightarrow \] ' = - 2
As Equation of normal PB, when slope is m,
y = mx - 2am -am3
Here m = m' = - 2 and a = 4
\[\therefore\,\,\,\] y = -2x - 2(4) (-2) - 4 . (-2)3
\[ \Rightarrow \] y = - 2x + 16 + 32
\[ \Rightarrow \] y = - 2x + 48 ..... (2)
At point B, y = 0
puttig y = 0 at equation (2) we get,
0 = - 2x + 48
\[ \Rightarrow \]x = 24
\[\therefore\,\,\,\] Coordinate of point B = (24, 0)
A circle is passing through point P, A and B, and C is the center of the circle.
So, AC and BC are the radius.
Then AC = BC, So C is the middle point of line AB.
\[\therefore\,\,\,\] C = \[\left( {{{24 - 16} \over 2},{{0 + 0} \over 2}} \right)\] = (4, 0)
\[\angle \]CPB = \[\theta \] and we have to find tan \[\theta \].
Slope of line PC = \[{{16 - 0} \over {16 - 4}}\] = \[{4 \over 3}\] =m1
and we know slope of line PB = -2 = m2
\[\therefore\,\,\,\] tan \[\theta \] = \[\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|\]
\[ \Rightarrow \] tan \[\theta \] = \[\left| {{{{4 \over 3} + 2} \over {1 - \left( {{4 \over 3}} \right)\left( 2 \right)}}} \right|\]
\[ \Rightarrow \] tan \[\theta \] = \[\left| {{{{{10} \over 3}} \over { - {5 \over 3}}}} \right|\]
\[ \Rightarrow \] tan \[\theta \] = \[\left| { - 2} \right|\]
\[ \Rightarrow \] tan \[\theta \] = 2