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Find the equation of the tangent plane at the point $(1,1,1)$  to the conicoid $3x^2-y^2=2z$.
asked Nov 11, 2017 by randomisation (1,920 points)  

2 Answers

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

$ \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 randomisation (1,920 points)  
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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

Answer     3x-y-z=1
answered May 27 by rishi95verma (140 points)  
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