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Find the equation of the tangent plane and normal line to the surface x^2/2-y^2/3=z at (2,3,-1)

Find the equation of the tangent plane and normal line to the surface x^2/2-y^2/3=z at (2,3,-1)

Grade:12

1 Answers

Arun
25750 Points
5 years ago
Note that the tangent plane to the surface f(x,y,z) = k at (x0,y0,z0) is
z(x0,y0,z0)(x-x0) + fy(x0,y0,z0)(y-y0) + fz(x0,y0,z0)(z-z0) = 0.
fx(x,y,z) = ∂f(x,y,z)/∂x
= ∂(x2/2 - y2/3 - z)/∂x
= x
fy(x,y,z) = ∂f(x,y,z)/∂y
= - 2y/3
fz(x,y,z) = ∂f(x,y,z)/∂z
= -1
The gradient value at (2, 3, -1) is
∇f(2, 3, -1) = x(2, 3, -1), fy(2, 3, -1), fz(2, 3, -1)>
=
The tangent plane is given by
2(x - (2)) - 2(y - 3) -1 (z - (-1)) = 0
∴ 2x - 2y - z + 1= 0
 
Regards
Arun (askIITians forum expert)

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