Find the equation of the tangent plane at the point $(1,1,1)$  to the conicoid $3x^2-y^2=2z$.

$3x^2-y^2=2z \implies 3x^2-y^2-2z=0 \equiv f(x,y,z)$

$\frac{\mathrm{d} f}{\mathrm{d} x} = 6x = 6 at x = 1$

$\frac{\mathrm{d} f}{\mathrm{d} y} = -2y = -2 at y = 1$

$\frac{\mathrm{d} f}{\mathrm{d} z} = -2$

This gives $6x - 2y - 2z = 6-2-2 = 2$. Thus, the equation of the tangent plane at given point is $6x - 2y - 2z = 2$
answered Nov 11, 2017 by (1,920 points)
Given point lies on the curve then we applied T=0

x^2= xx1

y^2=yy1

2z=(z+z1)

given x1=y1=z1=1

T=0 is equation of tangent

3xx1-yy1-z-z1=0

put value

3x.1-y.1-z-1=0