Answer to Question #87014 in Analytic Geometry for RAKESH DEY

Question

The equations 2x^2+y^2+3z^2+4x+4y+18z+34=0, 2x^2-y^2=4y-4z-4x represent a real conic.
Whether the statement is true or false.
Give reasons for your answer.

Answer 1

Transform the equations:

1) 2x² + y² + 3z² + 4x + 4y + 18z + 34 = 0;

2(x + 1)² – 2 + (y + 2)² – 4 + 3(z + 3)² – 27 + 34 = 0;

2(x + 1)² + (y + 2)² + 3(z + 3)² + 1 = 0. (1)

2) 2x² - y² = 4y - 4z - 4x;

4z = -2x² + y² + 4y - 4x;

4z = (y + 2)² – 4 – 2(x + 1)² + 2;

4z = (y + 2)² –2(x + 1)² - 2;

z = 1/4 * (y + 2)² – 1/2 * (x + 1)² – 1/2. (2)

The first of these hasn’t any solution and the second is an equation of the saddle-shaped structure (known as a hyperbolic paraboloid or an anticlastic surface). Hence, the equations do not represent a real conic, and the statement is false.