Question

Find the nature of the planar section of the conicoid x²/3-y²/4=z by the plane
x+2y−z = 6

Answer 1

$\frac{x^2}{3}-\frac{y^2}{4}=z\\x+2y-z=6$

We will find $z$

$z=x+2y-6$

$\frac{x^2}{3}-\frac{y^2}{4}=x+2y-6\\$

multiplying by 12

$4x^2-3y^2-12x-24y=-72\\ 4(x^2-3x)-3(y^2+8y)=-72\\ 4(x-\frac{3}{2})^2-3(y+4)^2=-72+4\cdot\frac{9}{4}-3\cdot16\\ 4(x-\frac{3}{2})-3(y+4)^2=-111\\$

multiplying by $111$

$\frac{4(x-\frac{3}{2})}{111}-\frac{3(y+4)^2}{111}=-1\\ \frac{(x-\frac{3}{2})}{\frac{111}{4}}-\frac{(y+4)^2}{{\frac{111}{3}}}=-1\\$

hyperbola, conjugated hyperbola $\frac{(x-\frac{3}{2})}{\frac{111}{4}}-\frac{(y+4)^2}{{\frac{111}{3}}}=1\\$ ,

 center $(\frac{3}{2};-4)$ ,

semi-axes $a=\sqrt{\frac{111}{4}}, b=\sqrt{\frac{111}{3}}$ .