Find vertices of the skew quadrilateral formed by the four generators of the hyperboloid $x^2+4y^2-4z^2=196$ passing through $(10,5,1)$ and $(14,2,-2)$
Progress : I proceeded by writing a general equation of line passing from $(10,5,1)$ with Direction ratios $(l, m , n)$ and took any point on that line of form $(lr_1+10,mr_1+5,nr_1+1)$ and being a point on generator it should satisfy the equation of the given hyperboloid which gives a quadratic in $r_1$.
Not sure how to proceed after it to get the equation of the two generators from $(10,5,1)$ and similarly for $(14,2,-2)$.
Or is there a easier way to get the vertices without getting the generator equations. Any help appreciated.
Write $x^2 + 4 y^2 - 4 z^2 = 196 \,$ as,
$(x - 2 z)(x + 2z) = (14-2y)(14+2y)$
This gives us the below two set of equations -
$\frac{x - 2 z}{14-2y} = \frac{14+2y}{x + 2 z} = s$....(i)
$\frac{x - 2 z}{14+2y} = \frac{14-2y}{x + 2 z} = t$....(ii)
Solving for $x, y, z$ of intersection points of two systems of lines from (i) and (ii),
$\displaystyle x = \frac{14(1+st)}{s+t}, y = \frac{7(s-t)}{s+t}, z = \frac{7(1-st)}{s+t}$ ...(iii)
Instead of solving as above, you can also remember that intersection points of the generating lines are given by
$\displaystyle x = \frac{a(1+st)}{s+t}, y = \frac{b(s-t)}{s+t}, z = \frac{c(1-st)}{s+t}$ for hyperboloid of one sheet $\, \frac{x^2}{a^2} + \frac{y^2}{a^2} - \frac{z^2}{a^2} = 1$.
Plugging point $(10,5,1)$ in (i) and (ii), we get $s = 2, t = \frac{1}{3}$
and point $(14,2,-2)$ gives us $s = \frac{9}{5}, t = 1$.
Plugging other two combinations of values of $s, t -> (2, 1) \, , (\frac{9}{5},\frac{1}{3})$ in (iii), we get coordinates of other two vertices of the skew quadrilateral formed by the generating lines of the hyperboloid and they are -
$(14,\frac{7}{3}, -\frac{7}{3})$ and $(\frac{21}{2},\frac{77}{16}, \frac{21}{16})$.