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=1ax2+by2+cz2=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=1ax2+by2+cz2=1 is a sphere with the same centre as that of the ellipsoid.

1 Answer
Mar 24, 2017

See below.

Explanation:

Calling E->f(x,y,z)=ax^2+by^2+cz^2-1=0Ef(x,y,z)=ax2+by2+cz21=0

If p_i = (x_i,y_i,z_i) in Epi=(xi,yi,zi)E then

ax_ix+by_iy+cz_iz=1axix+byiy+cziz=1 is a plane tangent to EE because has a common point and vec n_i = (ax_i,by_i,cz_i)ni=(axi,byi,czi) is normal to EE

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