Answer to Question #91551 in Analytic Geometry for Ra

Question

Find the equation of the normal of the paraboloid 3x²+4z²+4y=0 at the point (2,-4,1). Also find the point where this line again intersects the paraboloid.

Answer 1

F(x,y,z) = 3x²+4y+4z² = 0 is a paraboloid;

F_x = 6x, F_y = 4, F_z = 8z are partial derivatives.

M=(2,-4,1) is the point.

The equation of normal is

(x-x₀)/F_x(M) = (y-y₀)/F_y(M) = (z-z₀)/F_z(M).

In our case,

(x-2)/(6*2) = (y+4)/4 = (z-1)/8

(x-2)/12 = (y+4)/4 = (z-1)/8.

In order to find the other point of intersection we need to create a parameter t:

(x-2)/12 = (y+4)/4 = (z-1)/8 = t

From that equation we get

x = 12t+2, y = 4t-4, z = 8t+1.

We put these values into the paraboloid's equation and get

3(12t+2)²+4(8t+1)²+4(4t-4) =

3(144t²+48t+4)+256t²+64t+4+16t-16=432t²+144t+12+256t²+64t+4+16t-16 =

688*t²+208t = 0.

t(688t+208) = 0.

So either t=0 (and in this case the point would be the one we already have), either t = -208/688 = -13/43.

For t = -13/43 we have:

x = 12*(-13/43)+2 = -156/43+2 = -70/43;

y = 4*(-13/43)-4 = 4*(-13/43-1) = 4*(-56/43) = -224/43;

z = 8*(-13/43)+1 = -104/43+1 = -61/43.

Answer: (-70/43, -224/43, -61/43)