題目應(yīng)該是求點M的軌跡方程
解;
OA⊥OB
設(shè)直線OA:y=kx,直線OB:y=-x/k
解下方程組:
y=kx
y^2=4px
得A(4P/K^2,4P/K)
同理,解下方程組:
y=-x/k
y^2=4px
得B(4PK^2,-4PK)
直線AB的斜率：kAB=K/(1-K^2)
OM⊥AB,kOM=-(1-K^2)/K
設(shè)M(X,Y) ,則
Y/X=-(1-K^2)/K,-X/Y=K/(1-K^2),K^2=(X+YK)/X,
直線AB：Y+4PK=[K/（1-K^2) ]*(X-4PK^2)
Y+4PK=(-X/Y)*(X-4PK^2)
X^2+Y^2+4PKY=4PXK^2=4PX*(X+YK)/X=4PX+4PKY
X^2+Y^2-4PX=0
(X-2P)^2+Y^2=(2P)^2
點M的軌跡方程為一個園,半徑=2P,園心坐標為(2P,0)