find two straight lines that lies on the hyperbolic paraboloid?

Question

Find the parametric form of the two straight lines that lies entirely on the hyperbolic paraboloid z=2x^2-y^2 and pass through the point (x,y,z)=(1,1,1)

Answer 1

A hyperbolic paraboloid is an example of a quadric ruled surface. The hyperbolic paraboloid with equation:

z = x²/a² - y²/b²

can be represented by the parametric equations:

x(s, t) = a(s + t)
y(s, t) = b(s - t)
z(s, t) = 4st

For the equation at hand:

z = 2x² - y²

a² = 1/2
b² = 1

a = 1/√2
b = 1

For the point (1, 1, 1) we have:

x(s, t) = (s + t)/√2 = 1
y(s, t) = s - t = 1
z(s, t) = 4st = 1

Solving for s and t we obtain:

s = (1 + √2)/2
t = (-1 + √2)/2

From here you can get one line by freezing s at the value above and varying t. And you can get the other line by freezing t at the value above and varying s.

We need two more points, one to create each line.
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First line L1.

To create the first point P1:

s = (1 + √2)/2
t = 0

x(s, t) = (s + t)/√2 = s/√2 = (1 + √2) / (2√2)
y(s, t) = s - t = s = (1 + √2)/2
z(s, t) = 4st = 4s*0 = 0

P1 = [(1 + √2) / (2√2), (1 + √2)/2, 0]

We have the directional vector u for the line L1:

u = <(1 + √2) / (2√2) - 1, (1 + √2)/2 - 1, 0 - 1>
u = <(1 - √2) / (2√2), (-1 + √2)/2, -1>

L1 = <1, 1, 1> + ku
L1 = <1, 1, 1> + k<(1 - √2) / (2√2), (-1 + √2)/2, -1>
where k is a scalar ranging over the real numbers
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Second Line L2.

To create the second point P2:

s = 0
t = (-1 + √2)/2

x(s, t) = (s + t)/√2 = t/√2 = (-1 + √2) / (2√2)
y(s, t) = s - t = t = (-1 + √2)/2
z(s, t) = 4st = 4*0*t = 0

P2 = [(-1 + √2) / (2√2), (-1 + √2)/2, 0]

We have the directional vector v for the line L2:

v = <(-1 + √2) / (2√2) - 1, (-1 + √2)/2 - 1, 0 - 1>
v = <(-1 - √2) / (2√2), (-3 + √2)/2, -1>

L2 = <1, 1, 1> + pv
L2 = <1, 1, 1> + p<(-1 - √2) / (2√2), (-3 + √2)/2, -1>
where p is a scalar ranging over the real numbers

Answer 2

Why, did you lose yours ?????