# Show that the path traced by the point of intersection of three mutual perpendicular tangent planes to the ellipsoid ax^2+by^2+cz^2=1 is a sphere with the same centre as that of the ellipsoid.?

## Show that the path traced by the point of intersection of three mutual perpendicular tangent planes to the ellipsoid $a {x}^{2} + b {y}^{2} + c {z}^{2} = 1$ is a sphere with the same centre as that of the ellipsoid.

Mar 24, 2017

See below.

#### Explanation:

Calling $E \to f \left(x , y , z\right) = a {x}^{2} + b {y}^{2} + c {z}^{2} - 1 = 0$

If ${p}_{i} = \left({x}_{i} , {y}_{i} , {z}_{i}\right) \in E$ then

$a {x}_{i} x + b {y}_{i} y + c {z}_{i} z = 1$ is a plane tangent to $E$ because has a common point and ${\vec{n}}_{i} = \left(a {x}_{i} , b {y}_{i} , c {z}_{i}\right)$ is normal to $E$

Let $\Pi \to \alpha x + \beta y + \gamma z = \delta$ be a general plane tangent to $E$ then

$\left\{\begin{matrix}{x}_{i} = \frac{\alpha}{a \delta} \\ {y}_{i} = \frac{\beta}{b \delta} \\ {z}_{i} = \frac{\gamma}{c \delta}\end{matrix}\right.$

but

$a {x}_{i}^{2} + b {y}_{i}^{2} + c {z}_{i}^{2} = 1$ so

${\alpha}^{2} / a + {\beta}^{2} / b + {\gamma}^{2} / c = {\delta}^{2}$ and the generic tangent plane equation is

$\alpha x + \beta y + \gamma z = \pm \sqrt{{\alpha}^{2} / a + {\beta}^{2} / b + {\gamma}^{2} / c}$

Now given three orthogonal planes

${\Pi}_{i} \to {\alpha}_{i} x + {\beta}_{i} y + {\gamma}_{i} z = {\delta}_{i}$

and calling ${\vec{v}}_{i} = \left({\alpha}_{i} , {\beta}_{i} , {\gamma}_{i}\right)$ and making

$V = \left(\begin{matrix}{\vec{v}}_{1} \\ {\vec{v}}_{2} \\ {\vec{v}}_{3}\end{matrix}\right)$ we can choose

$V \cdot {V}^{T} = {I}_{3}$

and as a consequence

${V}^{T} \cdot V = {I}_{3}$

then we have also

$\left\{\begin{matrix}{\sum}_{i} {\alpha}_{i}^{2} = 1 \\ {\sum}_{i} {\beta}_{i}^{2} = 1 \\ {\sum}_{i} {\gamma}_{i}^{2} = 1 \\ {\sum}_{i} {\alpha}_{i} {\beta}_{i} = 0 \\ {\sum}_{i} {\alpha}_{i} {\gamma}_{i} = 0 \\ {\sum}_{i} {\beta}_{i} {\gamma}_{i} = 0\end{matrix}\right.$

Now adding ${\sum}_{i} {\left({\alpha}_{i} x + {\beta}_{i} y + {\gamma}_{i} z\right)}^{2}$ we have

${x}^{2} {\sum}_{i} {\alpha}_{i}^{2} + {y}^{2} {\sum}_{i} {\beta}_{i}^{2} + {z}^{2} {\sum}_{i} {\gamma}_{i}^{2} + 2 \left(x y \sum \left({\alpha}_{i} {\beta}_{i}\right) + x z \sum \left({\alpha}_{i} {\gamma}_{i}\right) + \sum \left({\beta}_{i} {\gamma}_{i}\right)\right) = {\sum}_{i} {\delta}_{i}^{2}$

and finally

${x}^{2} + {y}^{2} + {z}^{2} = {\sum}_{i} {\delta}_{i}^{2}$

but ${\sum}_{i} {\delta}_{i}^{2} = {\sum}_{i} {\alpha}_{i}^{2} / a + {\sum}_{i} {\beta}_{i}^{2} / b + {\sum}_{i} {\gamma}_{i}^{2} / c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

so

${x}^{2} + {y}^{2} + {z}^{2} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

which is the path traced by the point of intersection of three mutual perpendicular tangent planes to the ellipsoid.

Attached a plot for the ellipsoid

${x}^{2} + 2 {y}^{2} + 3 {z}^{2} = 1$ 