Answer to Question #102912 in Analytic Geometry for ag

Question

Trace the conicoid represented by x^2 +2z^2 = y. Also describe its sections by the planes x = c,∀c ∈ R.

Answer 1

It follows from the Conicoid equation $y=x^2+2\cdot z^2$ that a solution exists only for $y>0$. For $y<0$ , there is no solution with $x,z\in R$ . The form of conicoid is displayed on fig.1. For $y>0$ , conicoid sections with planes $y=c\in R>0$ have the form of ellipses with the formula $\frac{x^2}{c}+\frac{ z^2}{c/2}=1$ (fig.2). The center of the ellipse is located on the axis Y. The small axis is directed along Z and has a length of $b=\sqrt{c/2}$. The large axis of the ellipses is directed along the X axis and has a length of $a=\sqrt{c}$ . The formula for this section in the new notations takes on a canonical form of ellipses $\frac{x^2}{a^2}+\frac{z^2}{b^2}=1$. The conicoid body takes on values within$-a<x<a; -b<z<b$.

The section by $x=c\in R$ has the equation $y=c^2+2\cdot z^2$ which is a parabola with an axis of symmetry parallel with axis Y. The canonical form of the parabola is $(y-c^2)=\frac{z^2}{2\cdot p}$ , where $p=1/4$, and $y=c^2$ is its vertex. The vertex moves out of plane XZ $(y=0)$ as c increases. All the sections are simalar to each other with the focus distance of parabola $F=\frac{p}{2}=1/8$. Due to symmetry about YZ plane for $c<0$ we get exactly the same cross sections.

fig.1

fig.2