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2 answers
Reflection through the origin takes (x,y) -> (-x,-y)
Since (-x)(-y) = xy = 1
the reflection of any point through (0,0) is also on the hyperbola -- the other branch, in fact.
Since A' is the image of A, it is also on the hyperbola.
Since (-x)(-y) = xy = 1
the reflection of any point through (0,0) is also on the hyperbola -- the other branch, in fact.
Since A' is the image of A, it is also on the hyperbola.
1 answer