Show that the path traced by the point of intersection of three mutual perpendicular tangent planes to the ellipsoid $a x^{2} + b y^{2} + c z^{2} = 1$ $ax^2+by^2+cz^2=1$ is a sphere with the same centre as that of the ellipsoid.?

Question

Show that the path traced by the point of intersection of three mutual perpendicular tangent planes to the ellipsoid $a x^{2} + b y^{2} + c z^{2} = 1$ $ax^2+by^2+cz^2=1$ is a sphere with the same centre as that of the ellipsoid.?

Show that the path traced by the point of intersection of three mutual perpendicular tangent planes to the ellipsoid $a x^{2} + b y^{2} + c z^{2} = 1$ $ax^2+by^2+cz^2=1$ is a sphere with the same centre as that of the ellipsoid.

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Answer 1 · 2017-03-24T23:35:14

See below.

Explanation:

Calling $E \to f (x, y, z) = a x^{2} + b y^{2} + c z^{2} - 1 = 0$ $E->f(x,y,z)=ax^2+by^2+cz^2-1=0$

If $p_{i} = (x_{i}, y_{i}, z_{i}) \in E$ $p_i = (x_i,y_i,z_i) in E$ then

$a x_{i} x + b y_{i} y + c z_{i} z = 1$ $ax_ix+by_iy+cz_iz=1$ is a plane tangent to $E$ $E$ because has a common point and ${\to n}_{i} = (a x_{i}, b y_{i}, c z_{i})$ $vec n_i = (ax_i,by_i,cz_i)$ is normal to $E$ $E$

Let $Pi->alpha x+beta y+gamma z=delta$ be a general plane tangent to $E$ then

${(x_i=alpha/(a delta)),(y_i=beta/(bdelta)),(z_i=gamma/(c delta)):}$

but

$ax_i^2+by_i^2+cz_i^2=1$ so

$alpha^2/a+beta^2/b+gamma^2/c=delta^2$ and the generic tangent plane equation is

$alpha x+beta y + gamma z=pmsqrt(alpha^2/a+beta^2/b+gamma^2/c)$

Now given three orthogonal planes

$Pi_i->alpha_i x+beta_i y+gamma_i z=delta_i$

and calling $vec v_i = (alpha_i ,beta_i ,gamma_i)$ and making

$V=((vec v_1),(vec v_2),(vec v_3))$ we can choose

$V cdot V^T = I_3$

and as a consequence

$V^Tcdot V = I_3$

then we have also

${(sum_i alpha_i^2 =1),(sum_i beta_i^2 =1),(sum_i gamma_i^2 =1),(sum_i alpha_i beta_i=0),(sum_i alpha_i gamma_i=0),(sum_i beta_i gamma_i =0):}$

Now adding $sum_i(alpha_i x+beta_iy+gamma_iz)^2$ we have

$x^2sum_i alpha_i^2+y^2sum_i beta_i^2+z^2sum_i gamma_i^2+2(xy sum(alpha_i beta_i)+xzsum(alpha_i gamma_i)+sum(beta_i gamma_i))=sum_i delta_i^2$

and finally

$x^2+y^2+z^2=sum_i delta_i^2$

but $sum_i delta_i^2=sum_ialpha_i^2/a+sum_ibeta_i^2/b+sum_igamma_i^2/c = 1/a+1/b+1/c$

so

$x^2+y^2+z^2=1/a+1/b+1/c$

which is the path traced by the point of intersection of three mutual perpendicular tangent planes to the ellipsoid.

Attached a plot for the ellipsoid

$x^2+2y^2+3z^2=1$

enter image source here

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Show that the path traced by the point of intersection of three mutual perpendicular tangent planes to the ellipsoid $a x^{2} + b y^{2} + c z^{2} = 1$ $ax^2+by^2+cz^2=1$ is a sphere with the same centre as that of the ellipsoid.?