May 26th: Confocal Quadrics (Existence & Orthogonality)

Goal: prove that at each point point in the n-dimensional Euclidean space, there passes precisely n confocal quadrics; they are mutually orthogonal.

Existence Proof:

We will prove this for n = 3; the proof can be generalized to any n.

Given a point $(x_0, y_0, z_0) \in \mathbb{R}^3$ . The family of confocal quadrics can be given by:

$\frac{x^2}{a - \lambda} + \frac{y^2}{b - \lambda} + \frac{z^2}{c - \lambda} = 1$ , for fixed $a, b, c \in \mathbb{R}$ .

To show that 3 distinct mutually confocal quadrics pass through $(x_0, y_0, z_0)$ , it suffices to show that $\lambda$ has 3 distinct roots in the equation:

$\frac{x_0^2}{a - \lambda} + \frac{y_0^2}{b - \lambda} + \frac{z_0^2}{c - \lambda} = 1$

Without loss of generality, suppose $a \leq b \leq c$ . How do we show that there are 3 distinct roots? We will not be computing any of the solutions using algebra. Instead, we will use the intermediate value theorem to show that the function $f(\lambda) = \frac{x_0^2}{a - \lambda} + \frac{y_0^2}{b - \lambda} + \frac{z_0^2}{c - \lambda}$ crosses $y = 1$ exactly 3 times.

Key observation: $f(\lambda)$ has 3 vertical asymptotes. We need to analyze the behaviour of the function to the left and to the right of each asymptote. We will then obtain:

Done!

Orthogonality: The n mutually confocal quadrics intersecting at a point are orthogonal to each other.

Again, we prove this for n = 3; generalization to any $n \in \mathbb{N}$ can be made.

Suppose

$\frac{x^2}{a - \lambda_1} + \frac{y^2]{b - \lambda_1} + \frac{z^2}{c - \lambda_1} = 1$ (*)

and

$\frac{x^2}{a - \lambda_2} + \frac{y^2]{b - \lambda_2} + \frac{z^2}{c - \lambda_2} = 1$ (**)

are 2 quadrics passing through the point $(x_0, y_0, z_0)$ .

Note that (*) is a level curve of $f(x, y, z) = \frac{x^2}{a - \lambda_1} + \frac{y^2]{b - \lambda_1} + \frac{z^2}{c - \lambda_1}$ and (**) is a level curve of $g(x, y, z) = \frac{x^2}{a - \lambda_2} + \frac{y^2]{b - \lambda_2} + \frac{z^2}{c - \lambda_2}$ .

It follows that $\nabla f(x, y, z)|_{(x_0, y_0, z_0)}$ is normal to the surface defined by (*) and $\nabla g(x, y, z)|_{(x_0, y_0, z_0)}$ is normal to the surface defined by (**) at $(x_0, y_0, z_0)$ .

Hence, to show that the two surfaces are orthogonal to each other, it suffices to show that the two corresponding gradient vectors are orthogonal.

$\nabla f(x, y, z)|_{(x_0, y_0, z_0)} = (\frac{2x_0}{a - \lambda_1}, \frac{2y_0}{b - \lambda_1}, \frac{2z_0}{c - \lambda_1})$ and

$\nabla g(x, y, z)|_{(x_0, y_0, z_0)} = (\frac{2x_0}{a - \lambda_2}, \frac{2y_0}{b - \lambda_2}, \frac{2z_0}{c - \lambda_2})$ .

Now, how do we show that $\nabla f(x, y, z)|_{(x_0, y_0, z_0)} \cdot \nabla g(x, y, z)|_{(x_0, y_0, z_0)} = 0$ ?

Key: $f(x_0, y_0, z_0) - g(x_0, y_0, z_0) = 0$

$\Leftrightarrow (\frac{x_0^2}{a - \lambda_1} + \frac{y_0^2]{b - \lambda_1} + \frac{z_0^2}{c - \lambda_1}) - (\frac{x_0^2}{a - \lambda_2} + \frac{y_0^2]{b - \lambda_2} + \frac{z_0^2}{c - \lambda_2}) = 0$

$\Leftrightarrow (\lambda_1 - \lambda_2)(\frac{x_0^2}{(a - \lambda_1)(a - \lambda_2)} + \frac{y_0^2}{(b - \lambda_1)(b - \lambda_2)} + \frac{z_0^2}{(c - \lambda_1)(c - \lambda_2)}) = 0$ (***).

By (***),

$\nabla f(x, y, z)|_{(x_0, y_0, z_0)} \cdot \nabla g(x, y, z)|_{(x_0, y_0, z_0)} = 2(\frac{x_0^2}{(a - \lambda_1)(a - \lambda_2)} + \frac{y_0^2}{(b - \lambda_1)(b - \lambda_2)} + \frac{z_0^2}{(c - \lambda_1)(c - \lambda_2)}) = 0$ .

This concludes the proof.

Something I’m currently thinking about: how do we construct these family of quadrics in the higher dimensions? For n = 2, given 2 foci, the ellipse and the hyperbola are formed by considering points for which the sum of the distances from the foci or the difference of such distances is a fixed value. What about for n = 3?

May 26th: Confocal Quadrics (Existence & Orthogonality)

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