The eccentricity of ellipse e: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is e, P is the point on e, from P to the circle x ^ 2 + y ^ 2 = B ^ 2, make tangent PA and Pb, a and B are tangent points, Ask if there is a point P, so that PA ⊥ Pb? If there is, calculate the point P coordinate; if not, explain the reason

The eccentricity of ellipse e: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is e, P is the point on e, from P to the circle x ^ 2 + y ^ 2 = B ^ 2, make tangent PA and Pb, a and B are tangent points, Ask if there is a point P, so that PA ⊥ Pb? If there is, calculate the point P coordinate; if not, explain the reason


Let P point coordinate (x1, Y1), the slope of PA and Pb be K and - 1 / K, the linear equations are y = KX + y1-kx1, y = - X / K + Y1 + X1 / K, and x ^ 2 + y ^ 2 = B ^ 2 to form the equations, tangent Δ = 0, the solutions are: B & # 178; + B & # 178; K & # 178; - (Y1) &# 178; + 2kx1y1-k & # 178; (x1) &# 178; = 0, B & # 178; + B & # 178; K & # 178