Question

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

Answer 1

$\dfrac{x^2}{3}-\dfrac{y^2}{4}=z$

$x+2y−z=6$

We will find z

$z=x+2y−6$

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

$4x^2-3y^2-12x-24y=-72\\ 4(x^2-3x)-3(y^2+8y)=-72\\$

$4(x-3)^2-3(y+2)^2=-72+9-48=-111\\$

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

This is a equation of conjugated hyperbola

$center (\frac{3}{2};-4)$

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