Show that what is the graph of each one of these equations.

Question

Given the following equations: $$1)\;\;x^{2}+2y^{2}+z^{2}-2x+4z-22=0$$ $$2)\;\;5x^{2}+6y^{2}+4z-4x=14$$ $$3)-x^{2}+y^{2}-z^{2}-2x+2z=0$$ $$4) x=z^2$$ Show that what is the graph of each one of these equations.

$$1)$$ $$x^{2}+2y^{2}+z^{2}-2x+4z-22=0$$ $$\frac{\left(x-1\right)^{2}}{27}+\frac{y^{2}}{\frac{27}{2}}+\frac{\left(z+2\right)^{2}}{27}=1$$

Which is an ellipsoid.

$$2)$$ $$5x^{2}+6y^{2}+4z-4x=14$$ $$5x^{2}+6y^{2}+4z-4x=14$$ $$\frac{x^{2}}{12}+\frac{y^{2}}{10}+\frac{z}{15}-\frac{x}{15}=\frac{14}{60}$$ $$\frac{5x^{2}-4x}{60}+\frac{y^{2}}{10}+\frac{z}{15}=\frac{14}{60}$$ $$\frac{\left(x-\frac{2}{5}\right)^{2}}{60}+\frac{y^{2}}{50}+\frac{z}{75}=\frac{\frac{14}{60}-\frac{4}{5\cdot60}}{5}$$

$$3)$$ $$-x^{2}+y^{2}-z^{2}-2x+2z=0$$ $$x^{2}-y^{2}+z^{2}+2x-2z=0$$ $$\left(x+1\right)^{2}-y^{2}+\left(z-1\right)^{2}=2$$

I don't know the last three cases,can someone help me?

Number 1) is not an oval because an oval is a 2D shape. Did you mean to say ellipsoid? — Toby Mak, Jun 28 '20 at 13:40
Your work does help you to identify the 3D objects (except for the last one). There are only a few 3D conic sections, so you can work out the answers through the process of elimination. If you don't know the names of these objects, search them up online or look through your textbook. — Toby Mak, Jun 28 '20 at 13:53
It may also help to start with the 2D cross-sections of the objects. For example, for question 3 if you let $z$ be a constant, then $(x+1)^2 - y^2$ equals some constant $C$, and this is a .... ? If you extend this shape to 3D by letting $z$ vary, you should be able to name the 3D object. — Toby Mak, Jun 28 '20 at 13:58
Rewrite $(3)$ as $\left(x+1\right)^{2}+\left(z-1\right)^{2}=2 +y^{2}$ so when $y = 0$ $(x,z)$ forms a circle centered at $(x,z) = (-1,1)$ with radius $\sqrt{2}$. As $y$ increases or decreases the circle gets bigger. Looking up my old text book its called a hyperboloid of one sheet — , Jun 28 '20 at 14:05
Rewrite $(2)$ as $\displaystyle \frac{\left(x-\frac{2}{5}\right)^{2}}{60}+\frac{y^{2}}{50}=\frac{\frac{14}{60}-\frac{4}{5\cdot60}}{5} - \frac{z}{75}$ , This forms a cone with an elliptic base in the $(x,y)$ plane and with a point when the right side $= 0$ — , Jun 28 '20 at 14:28
45465 - When the right side is zero its the point of the dome $(x,y,z) = (\frac{2}{5},0,\frac{33}{10})$ I graphed it, its a dome not a cone. — , Jun 29 '20 at 00:57

score 1 · Answer 1 · 2020-06-29T02:08:20.273

Equation $1$. Ellipsoid.

The skirt is due to the real hack in the code to deal with complex values. It shouldn't be there.

Equation $2$. Dome.

Equation $3$. Hyperboloid of one sheet. Note the y axis is vertical.

Octave:

figure 1;
tx = ty = [-5:0.1:5]';
[xx, yy] = meshgrid (tx, ty);
zplus = -2 + real(sqrt(27 - (xx-1).^2 -2*yy.^2)); # real is a hack
zminus = -2 - real(sqrt(27 - (xx-1).^2 -2*yy.^2)); # real is a hack
mesh (tx, ty, zplus);
hold on;
mesh (tx, ty, zminus);
xlabel ("x");
ylabel ("y");
zlabel ("z");
title("[1]: (x-1)^2/27 + 2y^2/27 + (z+2)^2/27 = 1");
figure 2;
tx = ty = [-5:0.1:5]';
[xx, yy] = meshgrid (tx, ty);
z = 75( (14/60 - 4/(560))/5 - (xx-2/5).^2/60 - yy.^2/50);
mesh (tx, ty, z);
xlabel ("x");
ylabel ("y");
zlabel ("z");
title("[2]: (x-2/5)^2/60 + y^2/50 + z/75 = 11/250");
figure 3;
tx = tz = [-5:0.1:5]';
[xx, zz] = meshgrid (tx, tz);
yplus = real(sqrt((xx+1).^2 + (zz-1).^2 - 2)); # real is a hack
mesh (tx, tz, yplus);
hold on;
mesh (tx, tz, -yplus);
xlabel ("x");
ylabel ("z");
zlabel ("y");
title("[3]: (x+1)^2 - y^2 + (z-1)^2 = 2");

45465 - There are many types of domes most are cut off sections of spheres or ellipsoids. Like an orange cut in half. Some links: https://mathworld.wolfram.com/TorisphericalDome.html https://www.copper.org/applications/architecture/arch_dhb/arch-details/domes_spires_vaults/dome_panel_layout.html — , Jun 29 '20 at 04:22
45465 - The $y$ and $z$ axes are swapped and the centers are shifted $(x+1)$ and $(z-1)$ but it still has the shape of a hyperbola of one sheet. The center and orientation do not change the class of the shape. A pyramid is still a pyramid no matter where it is positioned or its orientation. — , Jun 29 '20 at 08:14

Show that what is the graph of each one of these equations.

1 Answers1