Shortest Distance Between Parabola and Circle Center

To determine the correct options, we analyze the given circle and parabola step-by-step.

1. Circle Properties

The given equation of the circle is: $x^2 + y^2 - 4x - 16y + 64 = 0$

Rewriting it in standard form by completing the square: $(x - 2)^2 + (y - 8)^2 - 4 - 64 + 64 = 0 \implies (x - 2)^2 + (y - 8)^2 = 2^2$

Thus, the center $S$ and radius $R$ of the circle are: $\text{Center } S = (2, 8), \quad \text{Radius } R = 2$

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2. Finding Point $P$ on the Parabola

The given parabola is $y^2 = 4x$, which corresponds to $a = 1$.

A general point $P$ on this parabola in parametric form is given by: $P(t^2, 2t)$

The shortest distance between a given point $S$ and a smooth curve occurs along the normal to the curve at $P$. The equation of the normal to $y^2 = 4x$ at $P(t^2, 2t)$ is: $y = -tx + 2t + t^3$

Since this normal passes through the center $S(2, 8)$: $8 = -t(2) + 2t + t^3 \implies t^3 = 8 \implies t = 2$

Substituting $t = 2$ back into the coordinates of $P$: $P = (2^2, 2 \times 2) = (4, 4)$

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3. Evaluating Option (A)

The length of the line segment $SP$ is: $SP = \sqrt{(4 - 2)^2 + (4 - 8)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$

Thus, Option (A) is correct.

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4. Evaluating Option (B)

$Q$ is a point on the circle that divides $SP$ internally. Since $S$ is the center of the circle, the distance $SQ$ is equal to the radius of the circle: $SQ = R = 2$

The length $QP$ is: $QP = SP - SQ = 2\sqrt{5} - 2$

Now, calculating the ratio $SQ : QP$: $\frac{SQ}{QP} = \frac{2}{2\sqrt{5} - 2} = \frac{1}{\sqrt{5} - 1}$

Rationalizing the denominator: $\frac{1}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{\sqrt{5} + 1}{5 - 1} = \frac{\sqrt{5} + 1}{4}$

So, $SQ : QP = (\sqrt{5} + 1) : 4$.

Thus, Option (B) is incorrect.

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5. Evaluating Option (C)

The equation of the normal to the parabola at $P(4, 4)$ (where $t = 2$) is: $y = -2x + 2(2) + 2^3 \implies y = -2x + 12$

To find its $x$-intercept, set $y = 0$: $0 = -2x + 12 \implies 2x = 12 \implies x = 6$

Thus, the $x$-intercept of the normal is $6$.

Therefore, Option (C) is correct.

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6. Evaluating Option (D)

The slope of the line passing through $S(2, 8)$ and $P(4, 4)$ is: $m_{SP} = \frac{4 - 8}{4 - 2} = \frac{-4}{2} = -2$

Since $Q$ lies on the segment $SP$, the radius line $SQ$ lies along the line $SP$. The tangent to the circle at $Q$ is perpendicular to the radius line $SQ$ (and hence perpendicular to $SP$).

Thus, the slope of the tangent at $Q$ is: $m_{\text{tangent}} = -\frac{1}{m_{SP}} = -\frac{1}{-2} = \frac{1}{2}$

Therefore, Option (D) is correct.

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Final Conclusion

The correct options are A, C, and D.