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Let S be a surface given by the equation:

$$ x^2 - y^2 -z = 0 $$

and P be the point $(1,-1,0)$. Find the two lines contained in S that pass through P.

We're not looking for answers outright (hence the 'homework' tag) but hints towards attacking this problem would be greatly appreciated.

Edit: Thanks for all of your responses so far. We are having trouble understanding how to proceed:

We can produce a parametric representation of all lines passing through $(1,-1,0)$ in $R^2$, and substitute this into the surface:

$$ (1+ta)^2-(-1 + tb)^2 = tc $$

Expanding and reducing this we come to;

$$ t^2a^2 - t^2b^2 + 2ta + 2tb = tc $$ $$ ta^2 - tb^2 + 2a + 2b = c $$

In our notes, our lecturer appears to jump from this to

$$ 2a + 2b -c = a^2 - b^2 = 0 $$

Confusing as we have assumed $t=1$ and we have equated to 0, with no real explanation given. How is this done?

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Note that if all points of a line are to belong to the surface, the coefficients of all powers of the line parameter have to vanish individually when you substitute the line into the left-hand side. – joriki Aug 17 '11 at 17:48
up vote 1 down vote accepted

Implicit equation for surface $S$ can be written as $z = (x-y)(x+y)$. A line is given by $x(t) = x_0 + t d_x$, $y(t) = y_0 + t d_y$ and $z(t) = z_0 + t d_z$.

Check when lines will belong to $S$, and then impose that they pass through $P$.

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Hint: To get a start, note that $x^2-y^2=(x-y)(x+y)$. Think about planes $x+y=a$ or $x-y=b$.

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