<s>[INST] Draw the Ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ a tangent plane which cuts off equal segments on the coordinate axis. Draw the Ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ a tangent plane which cuts off equal segments on the coordinate axis.
First, I want to find the normal line of the plane which intersects the $x,y,z$ at some distance $h$ away from the origin.
$$<-h,h,0> \times<-h, 0, h> = <h^2, h^2, h^2> = h^2<1,1,1>$$
So I now need to find when there is a point where the gradient is pointing in that direction
$$\triangledown f = <\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2}>$$
and the point must also satisfy the ellipsoid equation. $x = ka^2/2, y = kb^2/2,z=kc^2/2$ plugging into the ellipsoid equation yields a $$k = \frac{2}{\sqrt{a^2+b^2+c^2}}$$ which makes the gradient $$\triangledown f = <\frac{a^2}{\sqrt{a^2+b^2+c^2}},\frac{b^2}{\sqrt{a^2+b^2+c^2}},\frac{c^2}{\sqrt{a^2+b^2+c^2}}>$$
knowing $k$ I can also find the $<x,y,z>$ position and use $\triangledown f \cdot<x-x_o,y-y_o,z-z_o> =0$. I get a horrific result of :
$$a^2x+b^2y+c^2z=\frac{a^4+b^4+c^4}{\sqrt{a^2+b^2+c^2}}$$
Which is not correct. Could someone aid in creating a solution?
 [/INST] Ellipsoid is given by, $ \displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
If a plane is tangent to ellipsoid at point $(x_1, y_1, z_1)$, its equation is given by,
$\displaystyle \frac{x x_1}{a^2} + \frac{y y_1}{b^2} + \frac{z z_1}{c^2} = 1$
If $d \gt 0$, we also know that the plane passes through three points with coordinates,  $(\pm d, 0, 0), (0, \pm d, 0)$ and $(0, 0, \pm d)$.
Now plugging them in the equation of the plane, we get
$x_1 = \pm \dfrac{a^2}{d}, y_1 = \pm \dfrac{b^2}{d}, z_1 = \pm \dfrac{c^2}{d}$
As $(x_1, y_1, z_1)$ is on the ellipsoid, from equation of the ellipsoid we get,
$d = \sqrt{a^2+b^2+c^2}$
That leads to equations of tangents being,
$\pm x \pm y \pm z = \sqrt{a^2+b^2+c^2}$
These are $8$ such planes, one each in each octant.
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