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    • Thread Starter

    Part B

    can someone explain plz ?
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    Sub the equation of the normal back into xy = c²:

    => t²x² + (c/t)x - (ct³)x = c²

    t³ + (c-ct4)x - tc² = 0

    Now use the quadratic equation to solve for x:

    x = {(ct4-c) ± √[(c-ct4)² + 4c²t4] } / 2t³

    If you simplify the bit in the root, you get: √[c²(t8+2t4+1)]

    = √[c²(t4+1)²]

    = ct4 + c

    So, x = [ct4-c + (ct4+c)]/2t³ or x = [ct4-c - (ct4+c)]/2t³

    After simplification,

    x = ct or x = -c/t³

    x = ct is for the point P, therefore x = -c/t³ is for the point Q.

    When x = -c/t³ , (-c/t³)y = c²

    y = -ct³

    Therefore, Q( -c/t³ , -ct³ )
    • Thread Starter

    thanks (bloody hell thats long ) do u know wat u need to get A grade on that paper ???? out of 75 ?????
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Updated: June 19, 2005
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