Sum of Intesections of Tangent Plane of Curve are Constant Example

Let  
\[x^{\frac{2}{3}}+y^{\frac{2}{3}}+z^{\frac{2}{3}}=a^{\frac{2}{3}}\]
  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  
\[x_1^2+y_1^2+z_1^2= a^{\frac{4}{3}}({x_0}^{\frac{2}{3}}+ {y_0}^{\frac{2}{3}}+ {z_0}^{\frac{2}{3}})= a^{\frac{2}{3}} a^{\frac{2}{3}}= a^{\frac{4}{3}}+ \]
 

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