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As shown in the figure, it is known that a straight line and a parabola y ^ 2 = 2px intersect with two points a and B, and OA is perpendicular to ob, OD is perpendicular to AB, and ab intersects with point D. the coordinates of point D are (2,1), and the value of P is obtained
∵OD⊥AB,kOD=1/2
∴kAB=-2
Let the line AB be y = - 2x + B
A(x1,y1),B(x2,y2)
Then Y1 & sup2; = 2px1, Y2 & sup2; = 2px2,
∴(y1y2)²=4p²x1x2,
∵OA⊥OB,kOA•kOB=-1
∴y1/x1• y2/x2= y1y2/ x1x2
=y1y2•y1y2/x1x2•y1y2=
(y1y2)²/ x1x2•y1y2
=4p²x1x2/ x1x2•y1y2
=4p²/ y1y2=-1
∴y1y2=-4p²,x1x2=4p²
Substituting y = - 2x + B into Y & sup2; = 2px
We obtain (- 2x + b) & sup2; = 2px
That is 4x & sup2; - (4B + 2P) x + B & sup2; = 0
∴x1x2= b²/4
∴b²/4=4p²,b²=16p²
Ψ B = 4P (take positive value)
The line AB is y = - 2x + 4P
Substituting point d (2,1) into
1=-2×2+4p
∴p=5/4
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