Finding the tangent plane and normal line

[Kork]

D is a set of all points (x,y) in R2 distinct from (0,0).
I have the funtion f: D --> R which is defined by:
f(x,y) = (2xy)/(x²+y²

Im asked to find the equations (in plural) of the tangelnt plane and the normal line to the graph of the function at (1,2)

My attempt:

I use the tangent plane z= f(a,b)+f₁(a,b)*(x-a) + f₂(a,b)*(y-b)

f(1,2) = 4/5
f₁(1,2) = 12/25
f₂(1,2) = -6/25

when I put that into the tangent plane equation and simplify I get: (12/25)x - (6/25)y+4/5

Is this my normal line?

Im very confused, because the assigment asks me to find multiple equations of the tangent plane which I have never done before.

Help is appreciated.

 

[HallsofIvy]

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D is a set of all points (x,y) in R2 distinct from (0,0).
I have the funtion f: D --> R which is defined by:
f(x,y) = (2xy)/(x²+y²

Im asked to find the equations (in plural) of the tangelnt plane and the normal line to the graph of the function at (1,2)

My attempt:

I use the tangent plane z= f(a,b)+f₁(a,b)*(x-a) + f₂(a,b)*(y-b)

f(1,2) = 4/5
f₁(1,2) = 12/25
f₂(1,2) = -6/25

when I put that into the tangent plane equation and simplify I get: (12/25)x - (6/25)y+4/5

Before, you had "z= " that. And what happened to the "- a" and "- b"?

Is this my normal line?

You just said it was the "tangent plane equation". A plane is not a line!

Im very confused, because the assigment asks me to find multiple equations of the tangent plane which I have never done before.

Help is appreciated.

Any equation can be written in a number of ways. For example, multiplying both sides by a constant will give a new equation for the same object. Or just move terms from one side to another.

 

[Kork]

So if I do like this:

z*2 = ((12/25)x - (6/25)y+4/5) * 2

it can be concidered a valid answer?