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# Calculus problem?

I am stuck on this math problem and I can’t seem to figure it out.

### 2 Answers

- jacob sLv 71 month ago
Given:

z=4(x-1)^2 +4(y+3)^2 + 1 , ( 2, -2, 9 )

fx= 4(x-1) , fy = 8(y + 3) , fz =1

at ( 2, -2, 9 )

So, eqn. of tangent is

=> fx(X-Xo) + fy(Y-Yo) + fz(Z-Zo) = 0

=> 4(x-2) + 8(y + 2) + 1(z-9) = 0

=> 4x-8 + 8y + 16 + z - 9 = 0

=> 4x +8y + z - 1= 0

=> 4x +8y + z = 1

=> z = -(4x +8y + 1)

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- idiot1Lv 41 month ago
Denote (2,-2,9) by P and let f(x,y)=4(x-1)^2+4(y+3)^2+1. Then the gradient of the function F(x,y,z) = z-f(x,y) at P is a normal at P to the surface z=f(x,y). The gradient of F has as components the partials therefore it equals (-f_x,-f_y,1) = (-8(x-1),-8(y+3),1). At point P this equals (-8,-8,1). Therefore the normal plane has the equation

-8(x-2)-8(y+2)+(z-9) = 0

or

z = 9+8(x-2)+8(y+2) = 8x+8y+9