Deborah Proctor

2022-10-17

Consider the surface $$z=(\frac{x}{2}{)}^{2}-(\frac{y}{3}{)}^{2}$$. Compute the tangent plane at each point. Find the point such that the normal vector is vertical at that point.

Plutbantonavv

Beginner2022-10-18Added 15 answers

Given surface

$$z=(\frac{x}{2}{)}^{2}-(\frac{y}{3}{)}^{2}$$

tangent plane at each point

Here $$f(x,y,z)=(\frac{x}{2}{)}^{2}+(\frac{y}{3}{)}^{2}-z$$

let a point in given surface, whose posiion vector is

$${r}_{0}=ai+bj+ck$$

let p be any point of tangent plane

$$(r-{r}_{0})\mathrm{\u25b3}f=0\phantom{\rule{0ex}{0ex}}\Rightarrow ((x-a)u+(y-b)j+(z-jk))\phantom{\rule{0ex}{0ex}}\Rightarrow (x-a)\frac{x}{2}+\frac{2(y-b)y}{9}-(z-c)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x(x-a)}{2}+\frac{2y}{9}(y-b)-(z-c)$$

This is the equation of tangent plane at each point.

$$z=(\frac{x}{2}{)}^{2}-(\frac{y}{3}{)}^{2}$$

tangent plane at each point

Here $$f(x,y,z)=(\frac{x}{2}{)}^{2}+(\frac{y}{3}{)}^{2}-z$$

let a point in given surface, whose posiion vector is

$${r}_{0}=ai+bj+ck$$

let p be any point of tangent plane

$$(r-{r}_{0})\mathrm{\u25b3}f=0\phantom{\rule{0ex}{0ex}}\Rightarrow ((x-a)u+(y-b)j+(z-jk))\phantom{\rule{0ex}{0ex}}\Rightarrow (x-a)\frac{x}{2}+\frac{2(y-b)y}{9}-(z-c)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x(x-a)}{2}+\frac{2y}{9}(y-b)-(z-c)$$

This is the equation of tangent plane at each point.

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