[Math] Finding equation of tangent plane to the surface $f(x, y, z)=x^2+2y^2-3z^2$ at P(2, -1, 1)

Question

$z-z_0=f_x()(x-x_0)+f_y()(y-y_0)$

$f_x=2x,f_y=4y$

$z-1=4(x-2)-4(y+1)$

$z=4x-4y-5$

Can someone tell me if this is right? Going over old problems I got wrong.

Answer 1

Remember that the gradient is always normal to the surface. Therefore, if you want a tangent plane at a point $\vec p$ you use the fact that the dot product of orthogonal vectors is zero to get: $$\nabla f \cdot (\vec {x}-\vec p)=0$$ As the equation of the plane tangent to surface $f$ at point $\vec p$ In your case, evaluate $$\nabla f=(2x,4y,-6z)$$ At $(2,-1,1)$ to get $(4,-4,-6)$ and plug into the above equation to get $$(4,-4,-6)\cdot (x-2,y+1,z-1)=0$$