一道九年级数学题,哪位大神能解答一下,谢谢!
1)、直线 k = OD/OB = (-2/3)/1 = -2/3,直线表达式 y = -2x/3 + 2/3;
点A、B代入抛物线方程,16a - 8 + c = 0,a + 2 + c = 0,联立,a = 2/3,c = -8/3;
抛物线表达式 y = 2x^2/3 + 2x - 8/3;
2)、在交点E,-2x/3 + 2/3 = 2x^2/3 + 2x - 8/3,x^2 + 4x - 5 = 0,x = -5,y = 4;
D为直角,PD⊥ED,直线 PD y = 3x/2 + 2/3,点P x = (-2/3)/(3/2) = -4/9,t1 = OP/1 = 4/9 s;
E为直角,PE⊥ED, 直线 PE y = 3x/2 + c,点E -15/2 + c = 4,c = 23/2,y = 3x/2 + 23/2;
点P x = (-23/2)/(3/2) = -23/3,t2 = 23/3 s;
P为直角,EP^2 + PD^2 = ED^2,
[ 4^2 + (x+5)^2 ] + [ (2/3)^2 + x^2 ] = [ ( 4-2/3)^2 + 5^2 ]
整理得 3x^2 + 15x + 8 = 0,x = (-5/2) ±√129/6;
t3 = ( 5/2 + √129/6 ) s;t4 = ( 5/2 - √129/6 ) s;
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