Find the equation of the tangent plane to the conicoid 4x^2 + z^2 = 3y at the point (0,1,√3)

Question

It is a question of analytical geometry

Answer 1

First, write F=4x^2-3y+z^2=0 as whole equation for the surface (level surface format), then we partially differentiate each term:

DF/Dx=8x; DF/Dy=-3; DF/Dz=2z where D is used to designate partial differential.

Plugging in (0,1,sqrt(3)) we get: DF/Dx=0, DF/Dy=-3; DF/Dz=2sqrt(3).

The normal is vector N=(0,-3,2sqrt(3)) but we need the equation of the tangential plane and for that we need vector r=(x,y,z) and the vector dot product (scalar product) of r.N=(0,1,sqrt(3)).N.

This gives us 0-3y+2zsqrt(3)=-3+6=3. The equation of the plane is 2zsqrt(3)-3y=3.

Answer 2

4x^2 + z^2 = 3y
point (0,1,√3)
d/dx = 8x
d/dy=-3
d/dz=2z

d/dx=0
d/dy=-3
d/dz=2√3
0x-3y+2z√3=3

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