# Find the planes tangent to the following surfaces at the

Find the planes tangent to the following surfaces at the indicated points:
a) ${x}^{2}+2{y}^{2}+3xz=10$, at the point ($1,2,\frac{1}{3}$)
b) ${y}^{2}-{x}^{2}=3$, at the point (1,2,8)
c) $xyz=1$, at the point (1,1,1)
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Malena

Let $f:{\mathbb{R}}^{n}\to \mathbb{R}$ be a differentiable function. Recal that the tangent plane of the surface consisting of points such that f(x)=k for some constant k at the point ${x}_{0}$ is given with:
$▽f\left({x}_{0}\right)\cdot \left(x-{x}_{0}\right)=0$ (1)
a) Here we set $f\left(x,y,z\right)={x}^{2}+2{y}^{2}+3xz,{x}_{0}=\left(1,2,\frac{1}{3}\right)$ and k=10. Note that $f\left({x}_{0}\right)=k$.
Now we can calculate:
$▽f\left(x\right)=\left(2x+3z,4y,3x\right)⇒▽f\left(1,2,\frac{1}{3}\right)=\left(3,8,3\right)$
Using (1) we easily get:
$0=\left(3,8,3\right)\cdot \left(x-1,y-2,z-\frac{1}{3}\right)=3\left(x-1\right)+8\left(y-2\right)+3\left(z-\frac{1}{3}\right)$
Thus, the tangent plane equation is:
$3x+8y+3z=20$
b) Here we set $f\left(x,y,z\right)={y}^{2}-{x}^{2},{x}_{0}=\left(1,2,8\right)$ and k=3. Note that $f\left({x}_{0}\right)=k$
Now we can calculate:
$▽f\left(x\right)=\left(-2x,2y,0\right)⇒▽f\left(1,2,8\right)=\left(-2,4,0\right)$
Using (1) we easily get:
$0=\left(-2,4,0\right)\cdot \left(x-1,y-2,z-8\right)=-2\left(x-1\right)+4\left(y-2\right)+0\left(z-8\right)$
Thus, the tangent plane equation is:
c) Here we set $f\left(x,y,z\right)=xyz,{x}_{0}=\left(1,1,1\right)$ and k=1. Note that $f\left({x}_{0}\right)=k$. Now we can calculate:
$▽f\left(x\right)=\left(yz,xz,xy\right)⇒▽f\left(1,1,1\right)=\left(1,1,1\right)$
Using (1) we easily get:
$0=\left(1,1,1\right)\cdot \left(x-1,y-1,z-1\right)=1\left(x-1\right)+1\left(y-1\right)+1\left(z-1\right)$
Thus, the tangent plane equation is: