Equation of the tangent plane to the ellipsoid

Question

I need help finding the equation of the tangent plane to the ellipsoid ${x^2\over{a^2}} + {y^2\over{b^2}} + {z^2\over{c^2}} = 1$, such that the sum of intercepts on the coordinate axes is minimized. I attempted to use the method of constrained extremum. I aimed to determine the minimal value of $2a + 2b + 2c$ under the constraint ${x^2\over{a^2}} + {y^2\over{b^2}} + {z^2\over{c^2}} = 1$, but I am not obtaining valid results. Could someone assist me in formulating the expression?

Answer 1

The answer has already been given, but here we present a slightly different approach and elaborate some of the steps. We have the ellipsoid:

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

Two vectors which describe the tangent plane at a point $(x, y, z)$ on the ellipsoid are:

$$ \textbf{r}_x = (1, 0, \partial_x z) $$ $$ \textbf{r}_y = (0, 1, \partial_y z) $$

with:

$$ \partial_x z = -\frac{x c^2}{z a^2} $$ $$ \partial_y z = -\frac{y c^2}{z b^2} $$

and so the plane tangent to the ellipsoid at $\textbf{r}_0 = (x, y, z)$ is:

$$ \textbf{r} = s \textbf{r}_x + t \textbf{r}_y + \textbf{r}_0 $$

the plane intercepts the $x$-axis at $\textbf{r} = (x_i, 0, 0)$, or:

$$ \begin{bmatrix}x_i \\0 \\ 0 \end{bmatrix} = s\begin{bmatrix}1 \\0 \\ \partial_x z \end{bmatrix} + t\begin{bmatrix}0 \\1 \\ \partial_y z \end{bmatrix} + \begin{bmatrix}x \\y \\z \end{bmatrix} $$

and so we get three equations:

$$x_i = s + x$$ $$0 = t + y$$ $$0 = s \partial_x z + t \partial_y z + z$$

Solving these three equations we get that the plane intercepts the $x$-axis at $x_i$:

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

Similarly we conclude the plane intercepts the $y$-axis and $z$-axis at $y_i$ and $z_i$, respectively:

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

We would like to minimise the sum of the intercepts $F(x, y, z) = x_i + y_i + z_i$:

$$ F(x, y, z) = x + y + z + \left( \frac{a^2}{ x} + \frac{b^2}{ y} \right) \frac{z^2}{c^2} + \left( \frac{a^2}{x} + \frac{c^2}{z} \right) \frac{y^2}{b^2} + \left( \frac{b^2}{y} + \frac{c^2}{z} \right) \frac{x^2}{a^2} $$

Simplifying we get:

$$ F(x, y, z) = x + y + z + \left( \frac{a^2}{ x} + \frac{b^2}{ y} \right) \left( 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} \right) + \left( \frac{a^2}{x} + \frac{c^2}{z} \right) \left( 1 - \frac{x^2}{a^2} - \frac{z^2}{c^2} \right) + \left( \frac{b^2}{y} + \frac{c^2}{z} \right) \left(1 - \frac{y^2}{b^2} - \frac{z^2}{c^2} \right) $$

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

$$ = \frac{c^2}{z} + 2\frac{a^2}{x} + \frac{b^2}{y} - x - \left( \frac{ z^2}{c^2} + \frac{y^2}{b^2}\right) \left( \frac{a^2}{x} \right) $$

$$ = \frac{c^2}{z} + 2\frac{a^2}{x} + \frac{b^2}{y} - x - \left( 1 - \frac{x^2}{a^2} \right) \left( \frac{a^2}{x} \right) $$

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

$$ F(x, y, z) = \frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z}$$

We are looking for the minimum of $F$ constrained to the surface of the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ which we can find using the method of Lagrange Multipliers. We simply state the result here and refer to the answer provided by @cr001 for the details of this step. The minimum sum of intercepts occurs at:

$$ x = \left(-\frac{a^4}{2 \lambda} \right)^{\frac{1}{3}} $$ $$ y = \left(-\frac{b^4}{2 \lambda} \right)^{\frac{1}{3}} $$ $$ z = \left(-\frac{c^4}{2 \lambda} \right)^{\frac{1}{3}} $$

with:

$$ \lambda = \left( \frac{a^{\frac{2}{3}} + b^{\frac{2}{3}} + c^{\frac{2}{3}}}{2^{\frac{2}{3}}} \right)^{\frac{3}{2}} $$

Answer 2

The tangent plane is perpendicular to vector $({2x_0\over a^2},{2y_0\over b^2},{2z_0\over c^2})$ at point $(x_0,y_0,z_0)$. Therefore the tangent plane (that passes through the above point) equation is ${2x_0\over a^2}(x-x_0)+{2y_0\over b^2}(y-y_0)+{2z_0\over c^2}(z-z_0)=0$.

The intersection with the $x$-axis is calculated by setting $y,z=0$ and solve for $x$, which is $x_0+{1\over x_0}{a^2\over b^2}{y_0}^2+{1\over x_0}{a^2\over c^2}{z_0}^2$. By symmetry we can find out the other two axis-intersections and the sum of the coordinates is $x_0+y_0+z_0+({b^2\over a^2y_0}+{c^2\over a^2z_0}){x_0}^2+({a^2\over b^2x_0}+{c^2\over b^2z_0}){y_0}^2+({a^2\over c^2x_0}+{b^2\over c^2y_0}){z_0}^2$.

This means we want to minimize the function

$$F=x+y+z+({b^2\over a^2y}+{c^2\over a^2z}){x}^2+({a^2\over b^2x}+{c^2\over b^2z}){y}^2+({a^2\over c^2x}+{b^2\over c^2y}){z}^2$$

$$=({a^2\over x} + {b^2\over y} + {c^2\over z})({x^2\over a^2}+{y^2\over b^2}+{z^2\over c^2})$$

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

with constrant ${x^2\over{a^2}} + {y^2\over{b^2}} + {z^2\over{c^2}} = 1$, which can be done using Lagrange Multiplier method.

$$-{a^2\over x^2}={2\lambda x\over a^2}$$ $$-{b^2\over y^2}={2\lambda y\over b^2}$$ $$-{c^2\over z^2}={2\lambda z\over c^2}$$

Solve them you get $$x=({-a^4\over 2\lambda})^{1\over 3}$$ $$y=({-b^4\over 2\lambda})^{1\over 3}$$ $$z=({-c^4\over 2\lambda})^{1\over 3}$$

, and sub them into ${x^2\over{a^2}} + {y^2\over{b^2}} + {z^2\over{c^2}} = 1$ you get $({1\over 2\lambda})^{2\over 3}(a^{2\over 3} + b^{2\over 3} + c^{2\over 3}) = 1$ and $\lambda = \pm{1\over 2}(a^{2\over 3} + b^{2\over 3} + c^{2\over 3})^{3 \over 2}$.

The minimum is achieved when we take the positive $\lambda$ so $x,y,z$ are negative, which gives

$$x=-{a^{4\over 3}\over \left(a^{2\over 3} + b^{2\over 3} + c^{2\over 3} \right)^{\frac{1}{2}}}$$ $$y=-{b^{4\over 3}\over \left(a^{2\over 3} + b^{2\over 3} + c^{2\over 3} \right)^{\frac{1}{2}}}$$ $$z=-{c^{4\over 3}\over \left(a^{2\over 3} + b^{2\over 3} + c^{2\over 3} \right)^{\frac{1}{2}}}$$

And $$ F = -a^{\frac{2}{3}} \left(a^{\frac{2}{3}} + b^{\frac{2}{3}} + c^{\frac{2}{3}}\right)^{\frac{1}{2}} - b^{\frac{2}{3}} \left(a^{\frac{2}{3}} + b^{\frac{2}{3}} + c^{\frac{2}{3}}\right)^{\frac{1}{2}} - c^{\frac{2}{3}} \left(a^{\frac{2}{3}} + b^{\frac{2}{3}} + c^{\frac{2}{3}}\right)^{\frac{1}{2}} $$

$$ F = -\left( a^{\frac{2}{3}} + b^{\frac{2}{3}} + c^{\frac{2}{3}} \right)^{\frac{3}{2}} $$