Find the equation of the enveloping cylinder of the sphere #x^2+y^2+z^2-2x+4y=1# with its lines parallel to #x/2=y/3,z=0#?

Question

Find the equation of the enveloping cylinder of the sphere
#x^2+y^2+z^2-2x+4y=1#
with its lines parallel to
#x/2=y/3,z=0#

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Answer 1 · 2017-03-17T15:27:30

See below.

The directrix is given by

#L_d-> p= p_0+lambda vec v#

with #p=(x,y,z)#, #p_0=(0,0,0)# and #vec v = (2,3,0)#

The sphere is given by

#S->norm(p_s-p_1)=r#

where

#p_s=(x_s,y_s,z_s)#
#p_1=(1,-2,0)#
#r=sqrt(5)#

Now considering

#L_c->p=p_s+lambda vec v# as a generic line pertaining to the cylindrical surface, substituting #p_s = p - lambda vec v# into #S# we have

#norm(p-p_1-lambda vec v)^2=r^2# or

#norm(p-p_1)^2-2lambda << p-p_1, vec v >> + lambda^2 norm(vec v)^2 = r^2#

solving for #lambda#

#lambda = (2<< p-p_1, vec v >>pm sqrt((2<< p-p_1, vec v >>)^2-4(norm(p-p_1)^2-r^2)))/(2 norm(vec v)^2)#

but #L_c# is tangent to #S# having only a common point so

#(2<< p-p_1, vec v >>)^2-4norm(vec v)^2(norm(p-p_1)^2-r^2)=0# or

#<< p-p_1, vec v >>^2-norm(vec v)^2(norm(p-p_1)^2-r^2)=0#

This is the cylindrical surface equation. After substituting numeric values

# 9 x^2 + 4 y^2 + 13 z^2 +28 y - 42 x - 12 x y=16#

Attached a plot of the resulting cylindrical surface.

enter image source here