Question

Find a point P(a,b,c) on the graph z=ln(4x^2+y^2) such that the tangent plane at that point is parallel to the plane 2x+2y-z=3

Answer 1

I already found the gradient to be <8x/(4x^2+y^2) , 2y/(4x^2+y^2), -1>

And I figured that you would have to set the gradient equal to a scalar times the normal of the plane (which is n=<2,2,-1>

so my thoughts are that both the partial derivatives of z have to equal 2...

But I can't find any values that would make that work nicely... Did I make a mistake or is the question just mean?

Answer 2

may have just found it...

Answer 3

https://www.wolframalpha.com/input/?i=8x+%2F+%284x%5E2+%2B+y%5E2%29+%3D+2%2C+2y+%2F+%284x%5E2+%2B+y%5E2%29+%3D+2

Answer 4

there we go.

ended up trying a different method and it worked out

Answer 5

Just in time for me to solve it myself :D thanks anyways haha