So k =-2...(3 points)
(2) When k=- 1, d=( 1, k)=( 1,-1), the straight line L: Y =-X+2...(4 points).
The equations of line L and curve γ are sorted by elimination at the same time, x2-8x+4=0, where △ > 0 … (6 points).
Let A(x 1, y 1) and B(x2, y2), then x 1+x2=8, x 1x2=4…(7 points).
So | ab | =1+k2 | x1-x2 | = 46 ... (9 points)
(3) Suppose there is such a real number a that the symmetrical point Q(x0, y0) of the point P(a, 0) with respect to the straight line L falls on the quasi-line of the curve γ,
According to the meaning of the question, k > 0 can be obtained, so the straight line l: y = kx+2, because k > 0,
The equation of straight line L and curve γ eliminates Y at the same time, and it is concluded that k2x2+4(k- 1)x+4=0.
The straight line L and the curve γ have one thing in common, so △ =16 (k-1) 2-16k2 ≥ 0, and the solution is k≤ 12, so 0 < k≤12 .. (65433
Point P(a, 0) is about the symmetrical point Q(x0, y0) of the straight line L, so y02 = k? x0+a2+2y0x0？ a=？ 1k…( 12 points)
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