Ellipsoid $E:\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, and plane $P:lx + my + nz = 0$.
Assuming that the intersection of $E$ and $P$ is denoted as $D$, and the coordinates of the point d on D are represented as $(x, y, z) = (p, q, r)$. Find the maximum of $p$.
First, transform the ellipsoid into a sphere, be defining
$u = \dfrac{x}{a}, v = \dfrac{y}{b} , w = \dfrac{z}{c} $
Now we have the ellipsoid transformed into the unit sphere
$ u^2 + v^2 + w^2 = 1 $
Apply the same transformation to the plane, its equation in terms of $u, v, w$ becomes
$ a l \ u + b m \ v + c n \ w = 0 $
Let $N = ( a l , b m , c n ) $
The intersection of the plane and unit sphere is a circle with radius $1$, and a normal vector that is equal to $N$. Now let $r$ be define as
$ r = \sqrt{ (al)^2 + (bm)^2 + (cn)^2 } $
Then the unit vector along $N$ is
$ \hat{n} = \dfrac{N}{r} $
Further, $\hat{n}$ can be parameterized in spherical coordinates as follows
$ \hat{n} = ( \sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) $
Associated with $\hat{n}$ are two standard unit vectors that orthogonal to it, they are
$ u_1 = ( \cos \theta \cos \phi, \cos \theta \sin \phi, - \sin \theta )$
$ u_2 = ( - \sin \phi, \cos \phi , 0 ) $
The equation of the circle of intersection between the unit sphere and the plane is
$(u, v, w) = \cos(t) u_1 + \sin(t) u_2 $
Thus,
$(u,v,w) = ( \cos(t) \cos \theta \cos \phi - \sin(t) \sin \phi, \cos(t) \cos \theta \sin \phi + \sin(t) \cos \phi, - \cos(t) \sin \theta ) $
Now $x = a u = a (\cos(t) \cos \theta \cos \phi - \sin(t) \sin \phi ) $
The maximum of $x$ is given by
$ x_{Max} = a \sqrt{ \cos^2 \theta \cos^2 \phi + \sin^2 \phi } $
$ = a \sqrt{ 1 - \sin^2 \theta \cos^2 \phi } $
Using the above relation between $N$ and $\theta$ and $\phi$ we obtain,
$ \sin^2 \theta = \dfrac{ (al)^2 + (bm)^2 }{ (al)^2 + (bm)^2 + (c n)^2 }$
Therefore,
$ tan \phi = \dfrac{ b m }{a l } $
Hence,
$ cos^2 \phi = \dfrac{ 1 }{1 + tan^2 \phi } = \dfrac{ (al)^2 }{ (al)^2 + (bm)^2 }$
Therefore,
$ x_{Max} = a \sqrt{ 1 - \dfrac{ (al)^2 }{ ( (al)^2 + (bm)^2 + (cn)^2 ) }}$
which can also written as,
$ x_{Max} = a \sqrt{ \dfrac{(bm)^2 + (cn)^2 }{ (al)^2 + (bm)^2 + (cn)^2 }}$