Orthogonal projection of an ellipsoid

Question

Suppose an ellipsoid given by $\{{(x,y,z)| {x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{9}}}=1\}$, find the area of the orthogonal projection of the ellipsoid on the plane ${2x+4y-5z=10}$.

What is the right direction to solve this question? Thanks in advance!

Answer 1

Given the ellipsoid

$$ C\to p'Mp=1,\ \ \ p = (x,y,z), \ \ \ M = \mbox{diag}\left({1,\frac 14, \frac 19}\right) $$

and the plane

$$ \Pi\to(p-p_0)\cdot\vec v,\ \ \ \vec v = (2,4,-5) $$

the line

$$ L\to p=p_1+\lambda \vec v $$

is by construction, orthogonal to $\Pi$. Now making

$$ (p_1+\lambda\vec v)'M(p_1+\lambda\vec v)=1 $$

and solving for $\lambda$ we have

$$ \lambda = \frac{-2p_1'M\vec v\pm 2\sqrt{(p_1'M\vec v)^2-(p_1'M p_1-1)(\vec v'M\vec v)}}{2\vec v'M\vec v} $$

but if $L$ is tangent to $C$ then

$$ (p_1'M\vec v)^2-(p_1'M p_1-1)(\vec v'M\vec v)=0 $$

now assuming that $p_1'$ is the generic tangency point, the sought projection is given by the solution

$$ (p_1'M\vec v)^2-(p_1'M p_1-1)(\vec v'M\vec v)=0\\ (p_1-p_0)'\cdot \vec v = 0 $$

for $p_1$

which eliminating $z$ gives

$$ 7028 x^2 + 1088 x y + 2837 y^2 = 9700 $$

this is the trace projection on the $z=0$ plane. Eliminating the $x y$ term with a convenient rotation we get

$$ (1973 + \sqrt{749929}) X^2 + (1973-\sqrt{749929}) Y^2 = 3880 $$

with main axes

$$ r_X=2 \sqrt{\frac{970}{1973+\sqrt{749929}}}\\ r_Y=2 \sqrt{\frac{970}{1973-\sqrt{749929}}} $$

so the projected area is given by

$$ S_p = \pi r_X r_Y = \frac{2 \sqrt{97} \pi }{9} $$

but

$$ S_p = S \left(\frac{\vec v}{||\vec v||}\cdot \hat e_z\right) $$

hence

$$ S = \frac{2 \sqrt{97} \pi }{45} $$

Attached the problem setup

In lightblue $(p_1'M\vec v)^2-(p_1'M p_1-1)(\vec v'M\vec v)=0$ in light red $(p_1-p_0)'\cdot \vec v = 0$ and in light yellow $7028 x^2 + 1088 x y + 2837 y^2 = 9700$

Answer 2

Without calculus: The outline of the image of a quadric $Q$ under the central projection with matrix $P$ is a conic with a simple formula expressed via dual conics/quadrics: $C^*=PQ^*P^T$. What this formula says is that the outline of the image of $Q$ is the conic that is tangent to the images of the tangent planes of $Q$ that pass through the center of projection.

If we apply a rotation $R$ that aligns the projection direction with the $z$-axis, our orthogonal projection will then have the matrix $$P = \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\end{bmatrix} R,$$ i.e., the projection consists of rotating and then dropping the $z$-coordinate. The first two rows of $P$ are therefore the first two rows of $R$, for which we can take any two orthogonal unit vectors perpendicular to the image plane’s normal. $(4,-2,0)^T$ is obviously one such vector, and we can obtain another via a cross product: $(2,4,-5)^T\times(4,-2,0)^T=(-10,-20,-20)^T$. Normalizing these two vectors gives us $$P=\begin{bmatrix}{2\over\sqrt5} & -{1\over\sqrt5} & 0 & 0 \\ -\frac13 & -\frac23 & -\frac23 & 0 \\ 0&0&0&1 \end{bmatrix}.$$ The matrix $Q = \operatorname{diag}(1,1/4,1/9,-1)$, so using $Q^{-1}$ for the dual conic $Q^*$ is convenient. Multiplying this all out produces $$C^* = \begin{bmatrix}\frac85 & {2\over\sqrt5} & 0 \\ {2\over\sqrt5} & \frac{53}9 &0 \\ 0&0&-1 \end{bmatrix}.$$ The inverse of this matrix represents an ellipse centered at the origin.

There’s no need to invert this matrix to obtain the equation of the ellipse, though, since the ellipse’s area can be computed directly from the dual matrix. The area of an ellipse is $\pi$ times the product of its semiaxis lengths, and the latter are the reciprocal square roots of the eigenvalues of the quadratic part of $C$. Therefore, the area of this ellipse is equal to $${\pi\over\sqrt{-\det C}} = \pi \sqrt{-\det C^*} = \frac23\sqrt{97\over5}\pi.$$

With calculus: One possible approach is to find a parameterization of the outline of the projection. The generator of the outline of a quadric surface—the locus of intersection of rays from the projection center tangent to the surface—it itself a conic that lies in the polar plane of the projection center. For this parallel projection the center is the point at infinity $\mathbf V=(2,4,-5,0)$, so the generator’s plane is $Q\mathbf V = (2,1,-5/9,0)^T$, i.e., $2x+y-\frac59z=0$. Since the ellipsoid is axis-aligned, a relatively easy way to find a parameterization for the intersection curve is to transform the surface to a unit sphere. This transformation changes the polar plane to $(2,2,-5/3,0)^T$. Proceeding as above, we find a pair of unit vectors orthogonal to this plane’s normal, leading to the circle $$\cos(t)\left(\frac1{\sqrt2},-\frac1{\sqrt2},0\right)^T+\sin(t)\left(-{5\over\sqrt{194}},-{5\over\sqrt{194}}-6\sqrt{2\over97}\right)^T.$$ Mapping this circle back to the original coordinate system produces $$\cos(t)\left(\frac1{\sqrt2},-\sqrt2,0\right)^T+\sin(t)\left(-{5\over\sqrt{194}},-5\sqrt{2\over97},-18\sqrt{2\over97}\right)^T.$$

Since we’re working with a parallel projection, we can use any plane parallel to the image plane, so we project this onto $2x+4y-5z=0$. This is equivalent to projecting onto the original image plane and then translating the center of the resulting ellipse to the origin. The projection onto this plane is easily computed via the orthogonal rejection from the plane’s normal, resulting finally in a parameterization $\mathbf r(t)$ for the outline of the ellipsoid’s projection. Its area is then equal to $\frac12\int_0^{2\pi}\|\mathbf r\times d\mathbf r\|$. You should find that the integrand simplifies nicely.

On the other hand, once we had the two unit vectors that generated the great circle of the unit sphere above, we could again proceed without calculus. Scaling and projecting these vectors produces the halves of a pair of conjugate diameters of the outline. A theorem of Appolonios states that the area of the triangle formed by them is equal to $\frac12ab$, and so the area of the ellipse is equal to $2\pi$ times this area. Calling these two vectors $\mathbf u$ and $\mathbf v$, the area of ellipsoid’s projection is therefore equal to $\pi\|\mathbf u\times\mathbf v\|$.