Let $P$ be the point on the parabola $y^{2} = 4x$ which is at the shortest distance from the center $S$ of the circle $x^{2} + y^{2} - 4x - 16y + 64 = 0$. Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then (1) $SP = 2\sqrt{5}$ (2) $SQ : QP = \left(\sqrt{5} + 1\right) : 2$ (3) the $x$-intercept of the normal to the parabola at $P$ is $6$ (4) the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$

Let P be the point on the parabola y² = 4x which is at the shortest distance from the center S of the circle x² + y² - 4x - 16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then

$y^2 = 4x \quad (\text{parabola})$

$x^2+y^2-4x-16y+64=0 \quad (\text{circle})$

$S = \text{centre of the circle}$

$P \in y^2=4x \text{ nearest to } S$

$Q \in \text{circle}, \ Q \text{ divides } SP \text{ internally}$

$\text{To find: correct statement(s)}$

$x^2+y^2-4x-16y+64=0$ $\Rightarrow (x-2)^2+(y-8)^2=4$ $\Rightarrow S(2,8),\quad r=2$ $y^2=4x \Rightarrow a=1,\ \text{vertex } (0,0)$ $1.\ \text{shortest } SP \Rightarrow SP \text{ is the normal at } P$ $2.\ SP \text{ through } S \Rightarrow P$ $3.\ \text{check options (1)--(4)}$ $SP \perp \text{tangent at } P \Rightarrow SP \text{ is the normal}$ $\text{normal at param } t:\ y = -t\,x + 2t + t^3 \ (a=1)$ $S(2,8):\ 8 = -2t + 2t + t^3 \Rightarrow t^3 = 8$ $\Rightarrow t = 2$ $P = (t^2, 2t) = (4,4)$ $SP = \sqrt{(4-2)^2+(4-8)^2} = \sqrt{20} = 2\sqrt{5} \ \checkmark$ $SQ = r = 2,\quad QP = 2\sqrt{5}-2$ $\dfrac{SQ}{QP} = \dfrac{2}{2\sqrt5-2} = \dfrac{1}{\sqrt5-1} = \dfrac{\sqrt5+1}{4} \neq \dfrac{\sqrt5+1}{2} \ \times$ $t=2:\ y = -2x+12$ $y=0 \Rightarrow x = 6 \ \checkmark$ $m_{SQ} = \dfrac{4-8}{4-2} = -2$ $m_{\text{tangent}} = -\dfrac{1}{-2} = \dfrac12 \ \checkmark$ $\text{True: (1), (3), (4)};\quad (2) \text{ is false}$ $\boxed{\text{Correct options: (1), (3), (4)}}$