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

We have system of equations (conicoid and plane):$\begin{cases} \frac{x^2}{3}-\frac{y^2}{4}=z \\ x+2y-z=6 \end{cases} \quad \begin{cases} \frac{x^2}{3}-\frac{y^2}{4}=z\\ z=x+2y-6 \end{cases} \quad \begin{cases} \frac{x^2}{3}-\frac{y^2}{4}=x+2y-6\\ z=x+2y-6 \end{cases}$

$\frac{x^2}{3}-\frac{y^2}{4}=x+2y-6, \quad (\frac{x^2}{3}-x+\frac{3}{4})-\frac{3}{4}-(\frac{y^2}{4}+2y+4)+4=-6$

$(\frac{x}{\sqrt{3}}-\frac{\sqrt{3}}{2})^2-(\frac{y}{2}+2)^2=-9.25$ - equation of the hyperbolic cylinder.

$\begin{cases} (\frac{x}{\sqrt{3}}-\frac{\sqrt{3}}{2})^2-(\frac{y}{2}+2)^2=-9.25 \\ z=x+2y-6 \end{cases}$

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.