the set of all points P(x,y,z), such that the sum of the distances from P to the two points (0,-3,0) and (0,3,0) equals 10, forms a quadric surface. Identify this surface and give its equation in the standard form (x^2/a^2)+(y^2/b^2)+(z^2/c^2)=1

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the set of all points P(x,y,z), such that the sum of the distances from P to the two points (0,-3,0) and (0,3,0) equals 10, forms a quadric surface. Identify this surface and give its equation in the standard form (x^2/a^2)+(y^2/b^2)+(z^2/c^2)=1

Mathematics
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i wanna see the solution too ^^, just post sumthing here for able to folo its progress
\[d_1^2=x^2+(y+3)^2+z^2=x^2+y^2+z^2+9+6y=w+6y\]\[d_2^2=x^2+(y-3)^2+z^2=x^2+y^2+z^2+9-6y=w-6y\]\[w=x^2+y^2+z^2+9\]\[d_1+d_2=10\]\[d_1^2+2d_1d_2+d_2^2=100\]\[4d_1^2d_2^2=(100-d_1^2-d_2^2)^2\]\[4(w+6y)(w-6y)=(100-w-6y-w+6y)^2\]\[4(w^2-36y^2)=(100-2w)^2\]\[4w^2-144y^2=10000-400w+4w^2\]\[400w-144y^2-10000=16(25w-9y^2-625)=0\]\[25w-9y^2-625=0\]\[25(x^2+y^2+z^2+9)-9y^2-625=0\]\[25x^2+16y^2+25z^2=400\]\[\frac{x^2}{16}+\frac{y^2}{25}+\frac{z^2}{16}=1\]\[\frac{x^2}{4^2}+\frac{y^2}{5^2}+\frac{z^2}{4^2}=1\quad-\mbox{ellipsoid}\]
em i have a little question over here, that form : X2/a2 +y2/b2 + x2/c2 can applied on other shape of graph too ? like|dw:1327413868275:dw| or it can only apply on closed graph like circle, square and...?

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ellipsoid: xy-plane and yz-plane projections are ellipse, xz-plane projection is circle.
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if it is x and y only, if i encounter this kind of graph: |dw:1327415499115:dw| will i get the center point with that method?

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