# Plot Z = X^2 + Y^2 Over The Interval

I need khổng lồ find all points on the surface where the normal vector passes through the point $(0,0,2)$.

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I started with $ abla F = langle -2x,2y,1 angle$

Then, since the normal vector needs to pass through $(0,0,2)$ & $(x,y,z)$ then

$(-2x,2y,1) = lambda(x,y,z-2)$

So I get $lambda = pm1over 2$

I over up getting $z=3over 2$ & $z=5over 2.$

When $z=5over 2$, it appears lớn overshoot the point. It also appears that valid points only occur along the line $y=0$, so is it safe to assume that because this came from the relationship lớn $y$ it can be thrown out?

Therefore I get $x = pm sqrtfrac 32$ and therefore $(sqrtfrac 32,0,3over 2)$ and $(-sqrtfrac 32,0,3over 2)$.

Looking at the graph I should also get the point where $z=0$ & I"m not sure where that point gets left out & if I"m missing any others.

Is this the correct approach?

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edited Apr 5, 2022 at 7:55

Dr. Sundar

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asked Apr 5, 2022 at 7:50

lobo1335lobo1335

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## 2 Answers 2

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$egingroup$

At the point $(x,y,z)$ the gradient (the normal vector) is

$ n = (2 x, - 2 y, -1) $

The equation of the normal line is

$P(t) = (x, y, z) + t (2 x, -2 y, -1) $

Setting this equal to $(0,0,2)$ results in

$ (1 + 2 t) x = 0 , (1 - 2 t) y = 0, z = 2+t$, and in addition we have

$ z = x^2 - y^2 $

If $x =0, y=0$ then we have the point $(0,0,0)$.

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If $x = 0, y e 0$ then $t = dfrac12,$ và $z = dfrac52 $ , however, $y^2 = x^2 - z $ which has no real solutions.

If $x e 0, y = 0 $ then $ t= -dfrac12, z = dfrac32 ,x^2 = z + y^2 $ , for which the solutions are $x = pm sqrtdfrac32 $

If $x e 0 , y e 0$ , then there is no value of $t$ that will satisfy first two equations. Hence no more solutions.

Hence, there are three points $(0,0,0), (sqrtdfrac32 , 0, dfrac32 ) , (-sqrtdfrac32 , 0, dfrac32 )$

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answered Apr 5, 2022 at 11:06

Hosam HajeerHosam Hajeer

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$egingroup$

A normal"s equation through $(x,y,x^2-y^2)$ is given by$$eginpmatrixx\y\x^2-y^2endpmatrix+tcdot F_x imes F_y(x,y)=eginpmatrixx\y\x^2-y^2endpmatrix+teginpmatrix-2x\2y\1endpmatrix.$$Equating to $(0,0,0)$ gives the solutions $(x,y,t)=(0,0,-2)$ & $(pmsqrt3/2,0,1/2)$.

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answered Apr 5, 2022 at 8:58

Michael HoppeMichael Hoppe

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