Question

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

Answer 1

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`