A line through any point on the curve $x^2-y^2=1 , z=0$ intersects two lines $y=x, z=1$ and $y=-x, z=-1$. Required is the locus of the line . Just to clarify(as I have seen people question), the above form of a line represents the two planes that intersect to form the line.
Here I couldn't understand what does the problem mean by the locus of a line . I have seen this Meaning of locus of a line but it is inconclusive. It would be very helpful if someone could explain the reasons behind the solution of this question. Thankyou.
Put a point on the first line,
$P_1 = (t, t, 1) $
And a point on the second line
$P_2 = (s, -s, -1) $
Find the line that joins $P_1$ and $P_2$ ,
$P(t) = (1 - r) P_1 + r P_2 = ( (1 - r) t + r s , (1 - r) t - s r , (1 - r) -r ) $
Set $z = 0 $, then
$1 - 2 r = 0$, so $ r = \dfrac{1}{2} $
So the intersection point is
$P(t,s,r) = ( \dfrac{1}{2} (t + s) , \dfrac{1}{2} (t - s) , 0 )$
Now we want $x^2 - y^2 = 1$ , hence
$ (t + s)^2 - (t - s)^2 = 4 $ or $ t s = 1 $
Substituting this,
$P(t, r) = ((1 - r) t + r / t, ( 1- r) t - r / t, 1 - 2 r ) $
which is the line
$ P(t, r) = (t ,t , 1) + r ( - t + \dfrac{1}{ t} , - t -\dfrac{ 1}{t} , -2 ) $
This is the locus of all lines satisfying the given conditions.