Answer to Question #102912 in Analytic Geometry for ag

Question #102912
Trace the conicoid represented by x^2 +2z^2 = y. Also describe its sections by the planes x = c,∀c ∈ R.
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Expert's answer
2020-02-18T05:41:21-0500

It follows from the Conicoid equation y=x2+2z2y=x^2+2\cdot z^2 that a solution exists only for y>0y>0. For y<0y<0 , there is no solution with x,zRx,z\in R . The form of conicoid is displayed on fig.1. For y>0y>0 , conicoid sections with planes y=cR>0y=c\in R>0 have the form of ellipses with the formula x2c+z2c/2=1\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=c/2b=\sqrt{c/2}. The large axis of the ellipses is directed along the X axis and has a length of a=ca=\sqrt{c} . The formula for this section in the new notations takes on a canonical form of ellipses x2a2+z2b2=1\frac{x^2}{a^2}+\frac{z^2}{b^2}=1. The conicoid body takes on values withina<x<a;b<z<b-a<x<a; -b<z<b.

The section by x=cRx=c\in R has the equation y=c2+2z2y=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 (yc2)=z22p(y-c^2)=\frac{z^2}{2\cdot p} , where p=1/4p=1/4, and y=c2y=c^2 is its vertex. The vertex moves out of plane XZ (y=0)(y=0) as c increases. All the sections are simalar to each other with the focus distance of parabola F=p2=1/8F=\frac{p}{2}=1/8. Due to symmetry about YZ plane for c<0c<0 we get exactly the same cross sections.

fig.1


fig.2





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