Question

The section of 2x^2+y^2=2(1-z^2) by the plane x+2=0 is a hyerbola.

True or false with full explanation

Answer 1

The general equation of a hyperbola is

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

Let us find the section of $2x^2+y^2=2(1-z^2)$ by the plane $x+2=0$:

$2(-2)^2+y^2=2(1-z^2)$

$8+y^2=2-2z^2$

$y^2+2z^2=-6$

Since the eqution has no real solution, the plane $x+2=0$ does not intersect $2x^2+y^2=2(1-z^2)$. The equation $y^2+2z^2=-6$ is not a hyperbola equation.

Answer: false