? asked in 科學及數學數學 · 9 years ago


1. P and Q are two variable points of parameters p and q on the parabola 8y=x^2 such that OP is perpendicular to OQ. O is the origin

(a). Find the equation of the chord PQ

(b). Show that the chord PQ passes through a fixed point

1 Answer

  • 9 years ago
    Favorite Answer

    P(4p, 2p² ), Q(4q, 2q² ) ≠ (0,0)

    vector OP dot OQ = 0 = 16pq+ 4(pq)² , pq= -4 , q= -4/p, Q( -16/p, 32/p²)

    slope of PQ = (p- 4/p)/2


    equation of line PQ : 2(y-2p²)= (p- 4/p)(x-4p) or 2y=(p-4/p)x+ 16

    parametric of quation of PQ= (4p- 4(p+4/p) t, 2p²+ (32/p²- 2p²) t ), 0 <= t <= 1

    Note: the midpoint of PQ= (x,y)=( 2p- 8/p, p²+ 16/p²)

    the equation of the midpoint of PQ is 4y= x² +32


    the chord PQ passes through a fixed point (0, 8)

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