# 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 ${x}^{2} / {a}_{1}^{2} + {y}^{2} / {b}_{1}^{2} = \frac{2 z}{c} _ 1$ ; ${x}^{2} / {a}_{2}^{2} + {y}^{2} / {b}_{2}^{2} = \frac{2 z}{c} _ 2$; ${x}^{2} / {a}_{3}^{2} + {y}^{2} / {b}_{3}^{2} = \frac{2 z}{c} _ 3$ 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 ${a}_{1} , {b}_{1} , {c}_{1} \in \mathbb{R} \setminus \left\{0\right\}$ above is a 3x3 column matrix

Mar 22, 2017

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

$\left\{\begin{matrix}0 + {y}_{0}^{2} / {b}_{1}^{2} = \frac{2 {z}_{0}}{c} _ 1 \\ 0 + {y}_{0}^{2} / {b}_{2}^{2} = \frac{2 {z}_{0}}{c} _ 2 \\ 0 + {y}_{0}^{2} / {b}_{3}^{2} = \frac{2 {z}_{0}}{c} _ 3\end{matrix}\right.$

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

$\left\{\begin{matrix}{x}_{0}^{2} / {a}_{1}^{2} + 0 = \frac{2 {z}_{0}}{c} _ 1 \\ {x}_{0}^{2} / {a}_{2}^{2} + 0 = \frac{2 {z}_{0}}{c} _ 2 \\ {x}_{0}^{2} / {a}_{3}^{2} + 0 = \frac{2 {z}_{0}}{c} _ 3\end{matrix}\right.$

resulting in

${y}_{0}^{2} = \frac{2 {z}_{0} {b}_{1}^{2}}{c} _ 1 = \frac{2 {z}_{0} {b}_{2}^{2}}{c} _ 2 = \frac{2 {z}_{0} {b}_{3}^{2}}{c} _ 3$

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

$\det \left(\begin{matrix}{a}_{1}^{2} & {a}_{2}^{2} & {a}_{3}^{2} \\ {b}_{1}^{2} & {b}_{2}^{2} & {b}_{3}^{2} \\ {c}_{1} & {c}_{2} & {c}_{3}\end{matrix}\right)$

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

$\det \left(\begin{matrix}{a}_{1}^{2} & {a}_{2}^{2} & {a}_{3}^{2} \\ \lambda {c}_{1} & \lambda {c}_{2} & \lambda {c}_{3} \\ {c}_{1} & {c}_{2} & {c}_{3}\end{matrix}\right) = 0$

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

$\det \left(\begin{matrix}\eta {c}_{1} & \eta {c}_{2} & \eta {c}_{3} \\ {b}_{1}^{2} & {b}_{2}^{2} & {b}_{3}^{2} \\ {c}_{1} & {c}_{2} & {c}_{3}\end{matrix}\right) = 0$