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I want to take the projection of this ellipsoid into x-y, y-z and z-x plane. How can I do it?

ContourPlot3D[200.456+2.340*10^10*x^2+7.99*10^7*y^2+y*(2.80*10^(-9)-1.1735*10^6*z) - 
29150.591*z+1.895*10^6*z^2+x*(-4.329*10^6-9.731*10^7*y+3.135*10^8*z)==1, 
{x,0.00008,0.00011},{y,-0.00011,0.00022},{z,-0.0015,0.0015}]
user64494
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Sahabub Jahedi
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2 Answers2

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3D ellipsoid surface:

L[x_, y_, z_] = 200.4 + 2.3*10^10 x^2 + 8*10^7 y^2 + y*(2.8*10^-9 - 1.17*10^6*z) - 
  29150.6*z + 1.9*10^6*z^2 + x*(-4.3*10^6 - 9.73*10^7*y + 3.13*10^8*z) == 1;

project onto the $yz$ plane:

fx[y_, z_] = Resolve[Exists[x, L[x, y, z]], Reals]
(*    -1.97537*10^-8 - 0.000113839 y + 1. y^2 + 1.35294*10^-6 z - 0.00635739 y z + 0.0104524 z^2 <= 0    *)

project onto the $xz$ plane:

fy[x_, z_] = Resolve[Exists[y, L[x, y, z]], Reals]
(*    8.68073*10^-9 - 0.000187197 x + 1. x^2 - 1.26905*10^-6 z + 0.0135952 x z + 0.0000825289 z^2 <= 0    *)

project onto the $xy$ plane:

fz[x_, y_] = Resolve[Exists[z, L[x, y, z]], Reals]
(*    8.66424*10^-9 - 0.000187837 x + 1. x^2 - 8.87824*10^-7 y - 0.00009189 x y + 0.00789566 y^2 <= 0    *)

plots:

{RegionPlot[fx[y, z], {y, -0.00011, 0.00022}, {z, -0.0015, 0.0015}],
 RegionPlot[fy[x, z], {x, 0.00008, 0.00011}, {z, -0.0015, 0.0015}], 
 RegionPlot[fz[x, y], {x, 0.00008, 0.00011}, {y, -0.00011, 0.00022}]} // GraphicsRow

enter image description here

Roman
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If I correctly understand it, the OP asked about the subset within ranges {x,0.00008,0.00011}, {y,-0.00011,0.00022}, {z,-0.0015,0.0015}, This subset can be treated as

r = ImplicitRegion[200.4 + 2.3*10^10 x^2 + 8*10^7 y^2 + y*(2.8*10^-9 - 1.17*10^6*z) - 
 29150.6*z + 1.9*10^6*z^2 + 
 x*(-4.3*10^6 - 9.73*10^7*y + 3.13*10^8*z) == 1 && x >= 0.00008 &&
x <= 0.00011 && y >= -0.00011 && y <= 0.00022 && z >= -0.0015 && z <= 0.0015, {x, y, z}]

One of the ways to see its projection on $xOy$ plane consists in the specific point view, namely,

Region[r, ViewPoint -> {0, 0, Infinity}]

enter image description here

The two other projections are obtained by

Region[r, ViewPoint -> {0, Infinity, 0}]

and

Region[r, ViewPoint -> {Infinity, 0, 0}]

You may improve the qualities of the plots on your own.

user64494
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  • A simple way is to rotate your ContourPlot3D[200.456+2.340*10^10*x^2+7.99*10^7*y^2+y*(2.80*10^(-9)-1.1735*10^6*z) - 29150.591*z+1.895*10^6*z^2+x*(-4.329*10^6-9.731*10^7*y+3.135*10^8*z)==1, {x,0.00008,0.00011},{y,-0.00011,0.00022},{z,-0.0015,0.0015}] to see it from the above and from the other sides. – user64494 Apr 12 '21 at 20:27
  • An improvement to my answer is Region[r, ViewPoint -> {0, 0, Infinity}, AspectRatio -> 1]. – user64494 Apr 12 '21 at 20:31