Question

2x²+3Y²+4z²-4x-12y-24z+49=0 solve

Answer 1

Answer:

Step-by-step explanatio

Answer; taking lhs

Multiply both side by 0

0=0

Lhs =rhs

Answer 2

Answer:

The points satisfying this equation form an ellipsoid centred at (1, 2, 3) and with semi-axes of length 1/√2, 1/√3 and 1/2.

To see this, complete the squares in x, y and z.  So

2 x² + 3 y² + 4 z² - 4 x - 12 y - 24 z + 49 = 0

=> 2 ( x² - 2 x )  +  3 ( y² - 4 y )  +  4 ( z² - 6 z )  =  - 49

=> 2 ( x - 1 )² + 3 ( y - 2 )² + 4 ( z - 3 )² = - 49 + 2 + 12 + 36 = 1

Since

x² / a²  +  y² / b²  +  z² / c² = 1

is an ellipsoid centred at the origin with semi-axes a, b and c, we can read off the centre of our ellipsoid as the shift ( 1, 2, 3 ) and the semi-axes as the sqare roots of the reciprocals of the coefficients 2, 3 and 4.