Find the equation of the plane that is tangent to $x = y² + z² - 2$ on point P(-1,1,0)

Question

Find the equation of the plane that is tangent to the plane x = y² + z² - 2 on the point P(-1,1,0).

I got the equation x = 2y - 3 out of $F_{x}\Delta x + F_{y}\Delta y + F_{z}\Delta z = 0 $.

However this is a line, not a plane.

Are my calculations just plain wrong? Or if I did the math right, then what does this mean? Why do I get a line instead of a plane?

Answer 1

Your result is correct. The equation $x=2y-3$ is satisfied for any value of $z$ in $3D-$space and represents a plane parallel to the $z$ axis.

Answer 2

it not a line equation, it means the normal vector of the tangent plane is $(F_x,F_y,F_z)=(x,-2y,-2z)|_{-1,1,0}=(-1,-2,0)$, so the equation of the tangent plane with using $$ F_x(x-x_0)+F_y(y-y_0)+F_z(z-z_0)=0$$ is$$-(x+1)-2(y-1)+0(z-0)=0 $$ where $z\in R$ ,which every value should be take.