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# Show that if the point $(a, b, c)$ lies on the hyperbolic paraboloid $z = y^2 - x^2$, then the lines with parametric equations $x = a + t, y = b + t, z = c + 2 (b - a) t$ and $x = a + t, y = b - t, z = c - 2 (b + a) t$ both lie entirely on this paraboloid. (This shows that the hyperbolic paraboloid is what is called a ruled surface; that is, it can be generated by the motion of a straight line. In fact, this exercise shows that through each point on the hyperbolic paraboloid there are two generating lines. The only other quadric surfaces that are ruled surfaces are cylinders, cones, and hyperboloids of one sheet.)

## Since both parametric equations are equal to each other then can we saythat both of them lie entirely on $z=y^{2}-x^{2}$

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##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

so being given the point, ABC, with an equation of Z equals y squared minus X squared. We can plug in the point, um, and we get that C is equal to B squared minus a squared. So now we have the Z equals y squared minus X squared. That's how we end up getting the C equals B squared minus a squared. So now we want to plug in the Parametric equation into the equation of the hyperbolic parable oId and solve for C. So we have X equals a plus t. We have that y equals B plus T and Z equals C plus two times B minus 80 then plugging in. Now that we're solving for C, we end up getting that C plus to be a T is equal to Z, but we can now write Z as B plus T squared minus a plus T squared. Then C is going to C plus two B T minus 2 80 is going to equal B squared plus two B t plus T squared minus a squared minus to a T minus t squared. And then we can combine like terms, and this is going to end up giving us a nice C equals B squared minus a squared. Well, that looks familiar because we found it up here as well, then plugging in a second Parametric equation. Um, we have X equals a plus t y equals B minus T and see again equals negative or sorry Z equals C minus two times be this time plus a team. We multiply everything out. We set Z equal to those other values. So now we have is a C minus to be T minus 2 80 equals B squared minus to B T plus T squared minus a squared minus 2 80 minus t squared. We subtract those things over, and once again, we end up getting C equals B squared minus a squared. So since both Parametric equations are equal to each other, then we can say that both of them lie entirely on Z equal to y squared minus X squared.

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##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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