We have system of equations (conicoid and plane):{3x2−4y2=zx+2y−z=6{3x2−4y2=zz=x+2y−6{3x2−4y2=x+2y−6z=x+2y−6
3x2−4y2=x+2y−6,(3x2−x+43)−43−(4y2+2y+4)+4=−6
(3x−23)2−(2y+2)2=−9.25 - equation of the hyperbolic cylinder.
{(3x−23)2−(2y+2)2=−9.25z=x+2y−6
Intersection of hyperbolic cylinder and plane can be:
• a line, 2 parallel lines, no points (if plane is perpendicular to z=0 )
• hyperbola (if if plane isn’t perpendicular to z=0 )
In our case, x+2y−z=6 isn’t perpendicular to z=0.
Therefore, it will be hyperbola.
Answer: hyperbola.
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