Question

Find the circular section of ellipsoid x2+2y2+6z2=28?

Answer 1

Answer:

Step-by-step explanation:

x² + 2y² + 6z² = 28 (Given)

Dividing the equation by 28 on both the sides

= x²/28 + 2y²/28 + 6z²/28 = 28/ 28

= x²/28 + y²/14 + z²/14/3  = 1

= x²/28 + y²/14 + z²/14/3 - 1 = 0

Thus, s +  У.u. v = 0

where s = 0, represents a sphere and u = 0,  v = 0 are the equation of two plains

= x² + y²  + z². / 28 + y² ( 1/14 - 1/28 ) + z² (3/14 - 1/28) = 0

The pair of plain is given by -

= y² ( 1/14 - 1/28 ) + z² (3/14 - 1/28) = 0

= x² + y²  + z² = 28  

Hence the ellipsoid one by the planes given by two circles each of radius √28