Let P be the point on the parabola $y^{2} = 4 x$ which is at the shortest distance from the center S of the circle $x^{2} + y^{2} - 4 x - 16 y + 64 = 0$ .Let Q be the point on the circle dividing the line segment SP internally .Then:

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$S P = 2 \sqrt{5}$

The x-intercept of the normal to the parabola at P is 6.

$S Q : Q P = \sqrt{5} + 1 : 2$

The slope of the tangent to the circle at Q is $\frac{1}{2}$

answer is A, C, D.

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Detailed Solution

Let a general point be $p (t^{2}, 2 t)$
The shortest distance falls along common normal.
Normal to the circle passes through the center
Equating slopes of normal, we have at p , $\frac{2 t - 8}{t^{2} - 2} = - t \Rightarrow 2 t - 8 = - t^{3} + 2 t \Rightarrow t = 2$
P=(4,4)
Equation of normal is : $y = - 2 x + 12$
Slope of tangent at Q = $\frac{1}{2}$
$\frac{S P}{Q P} = \frac{2}{2 \sqrt{5} - 2} = \frac{1}{\sqrt{5} - 1} = \frac{\sqrt{5} + 1}{4}$