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.

1 Answer
Mar 24, 2017

Answer:

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#

enter image source here