Question

How do you find an equation for the horizontal tangent plane to the surface `z=4(x-1)^2+3(y+1)^2`

Answer 1

The answer is: `z=0`.

Remember that an horizontal plane is tangent to a curve in the space in its points of maximum, minimum or saddle.

We can answer in two ways.

The first: this function is the equation of an elliptic paraboloid with concavity upwards. Since `z` is surely positive or zero (it's the sum of two quantity positive or zero), the minimum, the vertex, is where it is zero, and this happens where:

`x=1 and y=-1`,

so the vertex is `V(1,-1,0)` and so the plane requested is the floor of the 3-dimensional space:

`z=0`.

The second is really long, but it's the only way if you don't recognize the curve.

We have to calculate the two partial derivative:

`(delz)/(delx)=8(x-1)`

`(delz)/(dely)=6(y+1)`

The system of the two previous equations (put `=0`) gives us the stationary points.

`8(x-1)=0`
`6(y+1)=0`

so:

`x=1`
`y=-1`

and we can calculate the applicate `z=0`.

Now we have to find what type of stationary point `P(1,-1,0)` is.
To do that we have to calculate the four partial derivative of the second order:

`(del^2z)/(delx^2)=8`

`(del^2z)/(dely^2)=6`

`(del^2z)/(delxdely)=(del^2z)/(delydelx)=0`

And the Hessian matrix:

`8,0`
`0,6`

Since its determinant in `P` is positive (`48`) and `(del^2z)/(delx^2)` is positive (`8`) the point `P` is a local minimum, so there exists an horizontal tangent plane and, since it has to pass from `P` which `z=0`, its equation is `z=0`.