Question

左右顶点分别为A（-2,0）,B（2,0）,离心率e=√3/2

Answer 1

(I)

a = 2,e² = 3/4= c²/a² = c²/4,c² = 3,b² = a² - c² = 4 - 3 = 1

x²/4 + y² = 1

(II)

P(p,q),p²/4 + q² = 1 (1)C(x,y),x = p (2)

|QP| = |PC|,|CQ| = 2|QP|

|y| = 2|q|,q² = y²/4 (3)

将(2)(3)代入(1):x²/4 + y²/4 = 1

x² + y² = 4

(III)

C(u,v),u² + v² = 4

AC的斜率为k = (v - 0)/(u +2)= v/(u + 2)

AC的方程:y = [v/(u + 2)](x + 2)

取x = 2,y = 4v/(u + 2)

R(2,4v/(u + 2))

D(2,2v/(u + 2))

CD的斜率k' = [2v/(u + 2) - v]/(2 - u) = uv/(u² - 4) = uv/(-v²）= - u/v

OC的斜率k" = v/u

k'k" = (-u/v)(v/u) = -1

OC与CD相互垂直,即CD与E相切．