Intersection of surface and tangent plane

[dustbin]

Homework Statement

I have a surface given by z=x^2 - y^2 and its tangent plane at the point (x,y)=(1,1) given by z = 2x-2y. I am asked to compute the intersection of the tangent plane with the surface.

The Attempt at a Solution

I did the obvious and set x^2-y^2 = 2x -2y to find the x,y lying on their intersection. I moved all of the terms to the RHS and used the quadratic formula to obtain x=y and x+y=2.

If x=y, then obviously z=0. I'm a little lost on what to do with the x+y=2. I know what the two graphs look like and I know what their intersection looks like, but I am trying to justify my answer.

 

[Simon Bridge]

You have two equations and two unknowns - what would you normally do in that situation?

 

[Dick]

Homework Statement

I have a surface given by z=x^2 - y^2 and its tangent plane at the point (x,y)=(1,1) given by z = 2x-2y. I am asked to compute the intersection of the tangent plane with the surface.

The Attempt at a Solution

I did the obvious and set x^2-y^2 = 2x -2y to find the x,y lying on their intersection. I moved all of the terms to the RHS and used the quadratic formula to obtain x=y and x+y=2.

If x=y, then obviously z=0. I'm a little lost on what to do with the x+y=2. I know what the two graphs look like and I know what their intersection looks like, but I am trying to justify my answer.

Ok, so if x=y then z=0. x=y is a plane, z=0 is another plane. So that leads you to the solution that is the intersection of those two planes, which is the line (x,x,0) where x can have any value. Do something very similar with the x+y=2 solution. You'll get the that the solution is two lines, yes?

 

[dustbin]

Okay, if I take x^2-y^2=(x+y)(x-y)=z and plug in 2-x=y, I get 4-4y=z. If I do the same for 2-y=x, I get 4x-4=z. Adding these two equations together gives me 2x-2y=z. So if x+y=2, then 2x-2y=z. So we have the intersection as two lines formed by the intersections of x=y, z=0 and x+y=2, 2x-2y=z.

 

[Dick]

Okay, if I take x^2-y^2=(x+y)(x-y)=z and plug in 2-x=y, I get 4-4y=z. If I do the same for 2-y=x, I get 4x-4=z. Adding these two equations together gives me 2x-2y=z. So if x+y=2, then 2x-2y=z. So we have the intersection as two lines formed by the intersections of x=y, z=0 and x+y=2, 2x-2y=z.

Well, yes, ok. So the two lines of intersection are what? Describe them both.

 

[dustbin]

Sorry, I'm not 100% sure what you mean. Can we not describe them as the two lines formed by the intersection of the planes x=y and z=0 and the planes x+y=2 and 2x-2y=z?

We could describe one line by (x, x, 0) and the other by (x, 2-x, 4-4x), where x may be chosen freely for both.

Thank you for your help!

 

[Simon Bridge]

<puzzled>
Did I misread? According to post #1 the intersection satisfies:
(1) x=y
(2) x+y=2
(3) z=0

... these relations are repeated in post #4
... why not just sub (1) into (2)?

 

[Dick]

Sorry, I'm not 100% sure what you mean. Can we not describe them as the two lines formed by the intersection of the planes x=y and z=0 and the planes x+y=2 and 2x-2y=z?

We could describe one line by (x, x, 0) and the other by (x, 2-x, 4-4x), where x may be chosen freely for both.

Thank you for your help!

Sure, that's right.

 

[Dick]

<puzzled>
Did I misread? According to post #1 the intersection satisfies:
(1) x=y
(2) x+y=2
(3) z=0

... these relations are repeated in post #4
... why not just sub (1) into (2)?

Those relations come from factoring x^2-y^2=2x-2y. (x-y)(x+y)=2(x-y). The solutions are x-y=0 OR x+y=2. Not necessarily both at the same time.