22. (Total Score: 18 points) Subquestion 1 is worth 6 points, Subquestion 2 is worth 8 points, Subquestion 3 is worth 4 points.\\
Let $P _ { 1 } ( x _ { 1 } , y _ { 1 } )$, $P _ { 2 } ( x _ { 2 } , y _ { 2 } )$, $\cdots$, $P _ { n } ( x _ { n } , y _ { n } )$ ($n \geq 3$, $n \in \mathbb{N}$) be points on a conic section C, and let $a _ { 1 } = \left| OP _ { 1 } \right| ^ { 2
From $\left\{ \begin{array} { l } \frac { x ^ { 2 } } { 100 } + \frac { y ^ { 2 } } { 25 } = 1 \\ x _ { 3 } ^ { 2 } + y _ { 3 } ^ { 2 } = 70 \end{array} \right.$, we obtain $\left\{ \begin{array} { l } x _ { 3 } ^ { 2 } = 60 \\ y _ { 3 } ^ { 2 } = 10 \end{array} \right.$\\
$\therefore$ The coordinates of point $P_{3}$ can be $( 2 \sqrt { 15 } , \sqrt { 10 } )$.\\
(2) \textbf{Solution 1:} The minimum distance from the origin $O$ to each point on the conic curve $C : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ $(a > b > 0)$ is $b$, and the maximum distance is $a$. $\because a _ { 1 } = \left| O P _ { 1 } \right| ^ { 2 } = a ^ { 2 }$, $\therefore d < 0$, and $a _ { n } = \left| O P _ { n } \right| ^ { 2 } = a ^ { 2 } + ( n - 1 ) d \geq b ^ { 2 }$,\\
$\therefore \frac { b ^ { 2 } - a ^ { 2 } } { n - 1 } \leq d < 0$. $\because n \geq 3$, $\frac { n ( n - 1 ) } { 2 } > 0$, $\therefore S _ { n } = n a ^ { 2 } + \frac { n ( n - 1 ) } { 2 } d$ is increasing on $\left[ \frac { b ^ { 2 } - a ^ { 2 } } { n - 1 } , 0 \right)$,\\
thus the minimum value of $S _ { n }$ is $n a ^ { 2 } + \frac { n ( n - 1 ) } { 2 } \cdot \frac { b ^ { 2 } - a ^ { 2 } } { n - 1 } = \frac { n \left( a ^ { 2 } + b ^ { 2 } \right) } { 2 }$.\\
\textbf{Solution 2:} For each natural number $k$ $(2 \leq k \leq n)$,\\
from $\left\{ \begin{array} { l } x _ { k } ^ { 2 } + y _ { k } ^ { 2 } = a ^ { 2 } + ( k - 1 ) d \\ \frac { x _ { k } ^ { 2 } } { a ^ { 2 } } + \frac { y _ { k } ^ { 2 } } { b ^ { 2 } } = 1 \end{array} \right.$, we solve to get $y _ { k } ^ { 2 } = \frac { - b ^ { 2 } ( k - 1 ) d } { a ^ { 2 } - b ^ { 2 } }$\\
$\because 0 < y _ { k } ^ { 2 } \leq b ^ { 2 }$, we obtain $\frac { b ^ { 2 } - a ^ { 2 } } { k - 1 } \leq d < 0$, $\therefore \frac { b ^ { 2 } - a ^ { 2 } } { n - 1 } \leq d < 0$. The rest is the same as Solution 1.\\
(3) \textbf{Solution 1:} If the hyperbola $C: \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$, and point $P _ { 1 } ( a , 0 )$,\\
then for a given $n$, the necessary and sufficient condition for the existence of points $P _ { 1 } , P _ { 2 } , \cdots , P _ { n }$ is $d > 0$.\\
$\because$ The distance from the origin $O$ to each point on the hyperbola $C$ is $h \in [ |a| , +\infty )$, and $\left| O P _ { 1 } \right| = a ^ { 2 }$,\\
$\therefore$ Points $P _ { 1 } , P _ { 2 } , \cdots , P _ { n }$ exist if and only if $\left| O P _ { n } \right| ^ { 2 } > \left| O P _ { 1 } \right| ^ { 2 }$, i.e., $d > 0$.\\
\textbf{Solution 2:} If the parabola $C : y ^ { 2 } = 2px$, and point $P _ { 1 } ( 0 , 0 )$, then for a given $n$, the necessary and sufficient condition for the existence of points $P _ { 1 } , P _ { 2 } , \cdots , P _ { n }$ is $d > 0$. The reasoning is the same as above.\\
\textbf{Solution 3:} If the circle $C: ( x - a ) ^ { 2 } + y ^ { 2 } = a ^ { 2 }$ $(a \neq 0)$, and $P _ { 1 } ( 0 , 0 )$,\\
then for a given $n$, the necessary and sufficient condition for the existence of points $P _ { 1 } , P _ { 2 } , \cdots , P _ { n }$ is $0 < d \leq \frac { 4 a ^ { 2 } } { n - 1 }$.\\
$\because$ The minimum distance from the origin $O$ to each point on the circle $C$ is $0$, and the maximum distance is $2 |a|$,\\
and $\left| O P _ { 1 } \right| ^ { 2 } = 0$, $\therefore d > 0$ and $\left| O P _ { n } \right| ^ { 2 } = ( n - 1 ) d \leq 4 a ^ { 2 }$, i.e., $0 < d \leq \frac { 4 a ^ { 2 } } { n - 1 }$.