\[x^{\frac{2}{3}}+y^{\frac{2}{3}}+z^{\frac{2}{3}}=a^{\frac{2}{3}}\]
be a curve in \[\mathbb{R}{^3}\]
. The sum of the squares of the intercept of each tangent plane to the curve is constant.Let
\[F(x,y,z)= x^{\frac{2}{3}}+y^{\frac{2}{3}}+z^{\frac{2}{3}}-a^{\frac{2}{3}}=0\]
.The tangent planes are the solutions to
\[\nabla F( \mathbf{r} - \mathbf{r}_0 )= \frac{\partial F}{\partial x}|_{x_0,y_0,z_0)} (x-x_0)+\frac{\partial F}{\partial y} |_{x_0,y_0,z_0)} (y-y_0)+\frac{\partial F}{\partial z} |_{x_0,y_0,z_0)} (z-z_0)+\]
.\[\frac{\partial F}{\partial x}_{(x_0,y_0,z_0)} = (\frac{2}{3} x^{- \frac{1}{3}})_{(x_0,y_0,z_0)}= \frac{2}{3} {x_0}^{- \frac{1}{3}}\]
\[\frac{\partial F}{\partial y}_{(x_0,y_0,z_0)} = \frac{2}{3} y^{- \frac{1}{3}})_{(x_0,y_0,z_0)}= \frac{2}{3} {y_0}^{- \frac{1}{3}}\]
\[\frac{\partial F}{\partial z}_{(x_0,y_0,z_0)} = \frac{2}{3} z^{- \frac{1}{3}})_{(x_0,y_0,z_0)}= \frac{2}{3} {z_0}^{- \frac{1}{3}}\]
The equations of the tangent planes are
\[\frac{2}{3} {x_0}^{- \frac{1}{3}} (x-x_0)+ \frac{2}{3} {y_0}^{- \frac{1}{3}} (y-y_0)+ \frac{2}{3} {z_0}^{- \frac{1}{3}} (z-z_0)=0\]
which can be rearranged as \[x{x_0}^{- \frac{1}{3}}+y{y_0}^{- \frac{1}{3}} + z{z_0}^{- \frac{1}{3}}= {x_0}^{\frac{2}{3}} + {y_0}^{\frac{2}{3}} +{z_0}^{\frac{2}{3}} = a^{\frac{2}{3}}\]
.The intersection of the tangent plane with the
\[x\]
axis is \[x_1= a^{\frac{2}{3}}{x_0}^{\frac{1}{3}}\]
, the intersection with the \[y\]
axis is \[y_1= a^{\frac{2}{3}}{y_0}^{\frac{1}{3}}\]
and the intersection with the \[z\]
axis is \[z_1= a^{\frac{2}{3}}{y_0}^{\frac{1}{3}}\]
, and \[x_1^2+y_1^2+z_1^2= a^{\frac{4}{3}}({x_0}^{\frac{2}{3}}+ {y_0}^{\frac{2}{3}}+ {z_0}^{\frac{2}{3}})= a^{\frac{2}{3}} a^{\frac{2}{3}}= a^{\frac{4}{3}}+ \]