Where does the normal line to the paraboloid z = x2 + y2 at the point (1, 1, 2) intersect the paraboloid a second time?

the intersection point is ( -5/4, -5/4, -9/8)

Step-by-step explanation:

F(x,y,z) = x² +y² -z= 0

Then find differential of each terms then we have

∀f(x)=2x

∀f(y)=2y

∀f(z)=-1

The partial differential which is the director vector is at F(x,y,z)= (1,1,2)

Vf(1,1,2) = (2,2,-1)

But the given point is ( 1.1,2)

Then the parametric equation of normal line will be

x= 1+2t

y= 1+2t

z= 2 -t

The parametric equation can be formed as

(1+2t)² + (1+2t)² - (2-t)= 0

If we expand the expression above we have,

2(1+4t+4t²) - 2+t= 0

0= 8t² + 9t

t= 0 or t= -9/8

Substitute the value of t into the parametric equation above

At t=0

x= 1+2t; x= 1+2(0)=1

y= 1+2t; y= 1+2(0)=1

z=2 -t ; z= 2-(0)= 2

At =0 , we have (1,1,2)

At t= -9/8

x= 1+2t; x= 1+2(-9/8)=-5/4

y= 1+2t; y= 1+2(-9/8)=-5/4

z=2 -t ; z= 2-(-9/8)= -9/8

Therefore, the intersection point is ( -5/4, -5/4, -9/8)

a) 0,0; 1,1; and 2,2

step-by-step explanation:

these have the exact same coordinates, thus they are the correct function.