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?

Answer 1

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/(5*60))/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");