Take the ellipsoid for example $$(x^2/a^2)+(y^2/b^2)+(z^2/c^2)=1$$ in the x-y plane you have an ellipse described by $$(x^2/a^2)+(y^2/b^2)=1$$ (suppose z=constant) in the y-z plane you have an ellipse described by $$(y^2/b^2)+(z^2/c^2)=1$$ (x=constant) in the x-z plane you have an ellipse described by $$(z^2/c^2)+(x^2/a^2)=1$$ (y=constant)
I understand how the equation describing a 2d ellipse was derived and I'm pretty sure it can apply to an ellipse in the y-z or x-z plane but I don't understand how the equation describing the 3d ellipse was derived.
I thought about adding all three equations in the three planes but ended up with $$(2x^2/a^2)+(2y^2/b^2)+(2z^2/c^2)=3$$ Dividing by 3 $$(2/3)((x^2/a^2)+(y^2/b^2)+(z^2/c^2))=1$$ $$(2/3)(x^2/a^2)+(2/3)(y^2/b^2)+(2/3)(z^2/c^2)=1$$ Assuming my process is correct would the 2/3 be considered as part of; a on the first term $$(x^2/a^2)$$, b on the second term $$(y^2/b^2)$$ and c on the third term$$(z^2/c^2)$$? resulting in $$x^2/a^2+y^2/b^2+z^2/c^2=1?$$