Tiêu đề 60 Locus of a Point.pdf ✅

4a√(x − c) 2 + y 2 = 4a 2 − 4cx a 2 [(x − c) 2 + y 2 ] = a 4 − 2a 2 cx + c 2x 2 (c 2 − a 2 )x 2 c 2 − a 2y 2 = a 2 c 2 − a 4 x 2 a 2 − y 2 c 2 − a 2 = 1 Setting b 2 = c 2 − a 2 results in the equation of a hyperbola, x 2 a 2 − y 2 b 2 = 1