# Consider the surface z=(x/2)^2-(y/3)^2. Compute the tangent plane at each point. Find the point such that the normal vector is vertical at that point.

Consider the surface $z=\left(\frac{x}{2}{\right)}^{2}-\left(\frac{y}{3}{\right)}^{2}$. Compute the tangent plane at each point. Find the point such that the normal vector is vertical at that point.
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Plutbantonavv
Given surface
$z=\left(\frac{x}{2}{\right)}^{2}-\left(\frac{y}{3}{\right)}^{2}$
tangent plane at each point
Here $f\left(x,y,z\right)=\left(\frac{x}{2}{\right)}^{2}+\left(\frac{y}{3}{\right)}^{2}-z$
let a point in given surface, whose posiion vector is
${r}_{0}=ai+bj+ck$
let p be any point of tangent plane
$\left(r-{r}_{0}\right)\mathrm{△}f=0\phantom{\rule{0ex}{0ex}}⇒\left(\left(x-a\right)u+\left(y-b\right)j+\left(z-jk\right)\right)\phantom{\rule{0ex}{0ex}}⇒\left(x-a\right)\frac{x}{2}+\frac{2\left(y-b\right)y}{9}-\left(z-c\right)\phantom{\rule{0ex}{0ex}}⇒\frac{x\left(x-a\right)}{2}+\frac{2y}{9}\left(y-b\right)-\left(z-c\right)$
This is the equation of tangent plane at each point.