Question

已知椭圆x∧2╱a∧2＋y╱b∧2,长轴为4,右顶点为a,在椭圆上存在一点p,使op⊥pa,求短

Answer 1

P(x,y),A(2,0)

OP^2 = x^2+y^2,PA^2 = (x-2)^2 +y^2

OP^2 + PA^2 = OA^2 = 4

x^2+y^2 - 2x = 0

b^2/4 x^2 + y^2 = b^2

(1-b^2/4)x^2 - 2x + b^2 = 0

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