Question

Show that the path traced by the point of intersection of three mutual perpendicular tangent planes to the ellipsoid `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 `ax^2+by^2+cz^2=1` is a sphere with the same centre as that of the ellipsoid.

Answer 1

See below.

Explanation:

Calling `E->f(x,y,z)=ax^2+by^2+cz^2-1=0`

If `p_i = (x_i,y_i,z_i) in E` then

`ax_ix+by_iy+cz_iz=1` is a plane tangent to `E` because has a common point and `vec n_i = (ax_i,by_i,cz_i)` is normal to `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`