Prove that the paraboloids #x^2/a_1^2+y^2/b_1^2=(2z)/c_1# ; #x^2/a_2^2+y^2/b_2^2=(2z)/c_2#; #x^2/a_3^2+y^2/b_3^2=(2z)/c_3# Have a common tangent plane if?

Prove that the paraboloids #x^2/a_1^2+y^2/b_1^2=(2z)/c_1# ; #x^2/a_2^2+y^2/b_2^2=(2z)/c_2#; #x^2/a_3^2+y^2/b_3^2=(2z)/c_3#
Have a common tangent plane if
#|a_1^2 a_2^2 a_3^2|#
#|b_1^2 b_2^2 b_3^2|#
#|c_1^2 c_2^2 c_3^2|#
#=0#
Here #a_1, b_1, c_1 in RR \ {0}#

  • above is a 3x3 column matrix

1 Answer
Mar 22, 2017

Answer:

See below.

Explanation:

If the three paraboloids have a common tangent plane, this tangency can be attained only at #x=0# or at #y=0# or at #x=y=0# due to the fact that they are centered about the #z# axis and have #x=0# and #y=0# symmetries.

Considering now #z=z_0# if their tangency is attained over the #y# axis then

#{(0+y_0^2/b_1^2=(2z_0)/c_1),(0+y_0^2/b_2^2=(2z_0)/c_2),(0+y_0^2/b_3^2=(2z_0)/c_3):}#

or if the tangency is attained over the #x# axis,

#{(x_0^2/a_1^2+0=(2z_0)/c_1),(x_0^2/a_2^2+0=(2z_0)/c_2),(x_0^2/a_3^2+0=(2z_0)/c_3):}#

resulting in

#y_0^2=(2z_0b_1^2)/c_1=(2z_0b_2^2)/c_2=(2z_0b_3^2)/c_3#

or

#b_1^2/c_1=b_2^2/c_2=b_3^2/c_3=lambda#

Analogously we get at

#a_1^2/c_1=a_2^2/c_2=a_3^2/c_3= eta#

Considering now the determinant

#det((a_1^2,a_2^2,a_3^2),(b_1^2,b_2^2,b_3^2),(c_1,c_2,c_3))#

if tangency occurs, for instance over the #y# axis, the determinant will read

#det((a_1^2,a_2^2,a_3^2),(lambda c_1,lambda c_2,lambda c_3),(c_1,c_2,c_3))=0#

The same fact occurs if the tangency occurs over the #x# axis

#det((etac_1,eta c_2,eta c_3),(b_1^2,b_2^2,b_3^2),(c_1,c_2,c_3))=0#