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

1 Answer
Feb 21, 2015

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#.