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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.

goku
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1 Answers1

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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 $(\frac{21}{2},\frac{77}{16}, \frac{21}{16})$.

Math Lover
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  • understood, thanks! btw is there a general equation that works for all conicoid – goku Oct 13 '20 at 14:05
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    @goku you are welcome. Unfortunately there is no general equation that would work for them all. It will depend on the equation of the conicoid, as you saw we derived it from the equation of the hyperboloid. – Math Lover Oct 13 '20 at 14:14