shacha777
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设P(2cost,sint),
PM:y=kx-2kcost+sint,①
代入椭圆方程得(1/4+k^2)x^2+2k(sint-2kcost)x+(sint-2kcost)^2-1=0,
x1=2cost,
x2=[(sint-2kcost)^2-1]/[2cost(1/4+k^2)]
=[(4k^2-1)cost-4ksint]/(1/2+2k^2),
代入①,y2=[(4k^3-k)cost-4k^2*sint+(1/2+2k^2)(sint-2kcost)]/(1/2+2k^2)
=[-2kcost+(1/2-2k^2)sint]/(1/2+2k^2).
以-1/(4k)代k,得
x3={[1/(4k^2)-1]cost+1/k*sint}/[1/2+1/(8k^2)]
=[(1-4k^2)cost+4ksint]/(2k^2+1/2)=-x2,
y3={1/(2k)*cost+[1/2-1/(8k^2)]sint}/[1/2+1/(8k^2)]
=[2kcost+(2k^2-1/2)sint]/(2k^2+1/2)=-y2,
M(x2,y2),N(-x2,-y2),
∴MN恒过原点.
1年前
10