Question

Find the value of `k` such that `z=k` is a tangent plane to `z=x^2-4xy-2y^2+12x-12y-1`?

(Question Restore: portions of this question have been edited or deleted!)

Answer 1

`-31`

Explanation:

The equation

`z = k`

corresponds to a plane with normal vector given by

`vec n = (0,0,1)`

Now the normal to the surface

`S(x,y,z)=z-(x^2-4xy-2y^2+12x-12y-1)=0` is

`grad S = (-12 - 2 x+4y 12 +2y+ 4 y, 1)` and

`vec n = grad S` at

`{(-12 - 2 x+4y=0),(12 + 4 y+4x=0):}`

or at

`x = -4, y=1`

At this point we have

`z = 6^2-4*6*3-2*3^2-12*6+12*3-1=-31`

then the tangency point is

`(-4,1,-31)` and `k = -31`

Answer 2

`k =-31`

Explanation:

We have:

`z=x^2-4xy-2y^2+12x-12y-1` ..... [A]

First we rearrange the equation of the surface into the form `f(x,y,z)=0`, And so we define our surface function, `f`, by:

`f(x,y,z) = z-x^2+4xy+2y^2-12x+12y+1`

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

`bb(grad) f(x,y,z) = (partial f)/(partial x) bb(ul(hat(i))) + (partial f)/(partial y) bb(ul(hat(j))) + (partial f)/(partial z) bb(ul(hat(k)))`

remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

`bb(grad) f = (-2x+4y-12)bb(ul(hat(i))) + (4x+4y+12)bb(ul(hat(j))) + (1)bb(ul(hat(k)))`

So for any particular point `(a,b,c)` the normal vector to the surface is given by:

`bb(grad) f(a,b,c) = (-2a+4b-12)bb(ul(hat(i))) + (4a+4b+12)bb(ul(hat(j))) + bb(ul(hat(k)))`

If we consider the tangent plane equation `z=k`, then this has normal vector:

`bb(ul(n)) = bb(ul(hat(k)))`

We require that the two normal vectors coincide; thus:

`bb(grad) f(a,b,c) = +- bb(ul(n))`

Which means that:

`-2a+4b-12 = 0`
`\ \ \ 4a+4b+12 = 0`

Which if we solve simultaneously we get `a=-4` and `b=1`.

If we use the original surface equation [A] we can find `z` when `x=-4` and `y=1`, thus:

`z=16 + 16 - 2 - 48 - 12 - 1 = -31`

Indicating the that given plane is a tangent at the coordinate `(-4,1,-31)` with a normal vector, `bb(ul(hat(k)))`, which we can verify graphically: