If the axes are rectangular, then find the locus of the equal conjugate diameters of ellipsoid, x2/a2 + y2/b2 + - Maths - Conic Sections

Question

If the axes are rectangular, then find the locus of the equal
conjugate diameters of ellipsoid, x2/a2 +
y2/b2 + z2/c2 = 1

Manbar Singh · Accepted Answer

Let OP, OQ, OR be 3 equal conjugate semi diameters , thenOP2 + OQ2 + OR2 = a2 + b2 + c2Let OP2 = OQ2 = OR2 = r2Now, a2 + b2 + c2 = 3r2     
1Let xl = ym = zn = r, represent any such diameters where l2 + m2 + n2 = 1So, coordinates of the extremities of such diameters are lr, mr, nr
This point lies on the given ellipsoid, then     l2r2a2 + m2r2b2 + n2r2c2 = 1⇒l2a2 + m2b2 + n2c2 = 1r2⇒l2a2 + m2b2 + n2c2 = l2 + m2 + n2r2⇒l2a2 + m2b2 + n2c2 = 3l2 + m2 + n2 a2 + b2 + c2     Using 1So, any of the equal conjugate semi diameters is generator of the cone     x2a2 + y2b2 + z2c2 = 3x2 + y2 + z2 a2 + b2 + c2 ⇒x2a2 + b2 + c2a2 - 3 + y2a2 + b2 + c2b2 - 3  + z2a2 + b2 + c2c2 - 3  = 0⇒x2a22a2-b2-c2 + y2b22b2-c2-a2 + z2c22c2-a2-b2 = 0This is the required locus

If the axes are rectangular, then find the locus of the equal conjugate diameters of ellipsoid, x2/a2 + y2/b2 + z2/c2 = 1

If the axes are rectangular, then find the locus of the equal conjugate diameters of ellipsoid, x²/a² + y²/b² + z²/c² = 1