设抛物线y=ax^2(a为不等于0的常数)上两点P,Q的切线互相垂直,求:证明过P,Q的直线必过定点R
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证明(1)设点P(x1,ax1^2),点Q(x2,ax2^2) y'=2ax k1=2ax1,k2=2ax2 k1k2=4a^2x1x2=-1 ∴y-ax1^2=(ax2^2-ax1^2)(x-x1)/(x2-x1) y=a(x2+x1)(x-x1)+ax1^2=a(x2-x1)x-a(x2+x1)x1+ax1^2 =a(x2+x1)x-ax2x1=a(x2+x1)x-a[-1/(4a^2)]=a(x2+x1)x+1/(4a) ∴过P,Q的直线L必过定点R(0,1/(4a)) (2)PS方程:y-ax1^2=2ax1(x-x1)即y=2ax1x-ax1^2┈① QS方程:y=2ax2x-ax2^2┈② ①-②得2a(x1-x2)x=a(x1^2-x2^2) ∴x(s)=(x1+x2)/2 y(s)=2ax1(x1+x2)/2-ax1^2=ax1x2=-1/(4a) ∴两切线的交点S在定直线上y=-1/(4a)上
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