Answer to Question #105605 in Analytic Geometry for Akanksha

Question

Trace the conicoid represented by x^2+2z^2°y. Also describe its section by planes x=c, for all c belongs to R.

Answer 1

Trace the conicoid represented by $x^2+2z^2=y$ . Also describe its sections by

the planes $x=c,\forall c\in \R.$

It follows from the conicoid equation $y=x^2+2z^2$ that solution exists only for $y\gt 0$ .

For $y\lt 0,$ there is no solution with $x,z\in \R.$ The form of conicoid is displayed on fig.1 .

For $y\gt0,$ conicoid sections with planes $y=c\in\R\gt0$ have form of ellipses with the formul

$a\frac{x^2}{c}+\frac{z^2}{c/2}=1$ (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 direced

along the $X$ axis an has a length of $a=\sqrt{c}.$ The formula for this section in the new notations

take on a canonical form of ellipses $\frac{x^2}{a^2}+\frac{z^2}{b^2}=1$. The conicoid body takes on value within

$-a\lt x\lt a;-b\lt z\lt b.$

The section by $x=c\in \R$ has the equation $y=c^2+2z^2$ which is a parabola with an axis

os symmetry parallel ith 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 section are similar to each other with focus distance of

parabola $F=\frac{p}{2}=1/8.$ Due to symmetry about $YZ$ plane for $c\lt0$ we get exactly

the same cross sections.

fig.1

fig.2