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Yes it is. After central projection on the plane (Klein model for sphere) you obtain usual ellipse.

Also you can show it using triangle inequality. All proofs from euclidean plane works. For example this one: Suppose $F_1$ and $F_2$ foci of the ellipse. Take any two points $A$ and $B$ inside and reflect $F_2$ with respect to the line $AB$. New point denote by $F_2'$. Take any point $X$ on the segment $AB$. Suppose ray $F_1X$ intersect the segment $F_2'A$ (the case $F_2'B$ is the same) in the point $Y$. We have, $$F_1X+F_2X=F_1X+F_2'X< F_1X+XY+YF_2'=F_1Y+YF_2'< F_1A+AY+YF_2'=F_1A+AF_2'$$

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Yes it is. After central projection on the plane (Klein model for sphere) you obtain usual ellipse.

Also you can show it using triangle inequality. All proofs from euclidean plane works. For example this one: Suppose $F_1$ and $F_2$ foci of the ellipse. Take any to two points $A$ and $B$ inside and reflect $F_2$ with respect to the line $AB$. New point denote by $F_2'$. Take any point $X$ on the segment $AB$. Suppose ray $F_1X$ intersect the segment $F_2'A$ (case $F_2'B$ the same) in the point $Y$. We have, $$F_1X+F_2X=F_1X+F_2'X< F_1X+XY+YF_2'=F_1Y+YF_2'< F_1A+AY+YF_2'=F_1A+AF_2'$$

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Yes it is. After central projection on the plane (Klein model for sphere) you obtain usual ellipse.

Also you can show it using triangle inequality. All proofs from euclidean plane works. For example this one: Suppose $F_1$ and $F_2$ foci of the ellipse. Take any to points $A$ and $B$ inside and reflect $F_2$ with respect to the line $AB$. New point denote by $F_2'$. Take any point $X$ on the segment $AB$. Suppose ray $F_1X$ intersect the segment $F_2'A$ (case $F_2'B$ the same) in the point $Y$. We have, $$F_1X+F_2X=F_1X+F_2'X< F_1X+XY+YF_2'=F_1Y+YF_2'< F_1A+AY+YF_2'=F_1A+AF_2'$$