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26 June, 07:43

Find the equation of the tangent to the curve at x=3 for parametric equations

x = t + 1/t

y = t^2 + 1 / (t^2) when t is greater than 0

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  1. 26 June, 08:03
    0
    y = 6x - 11

    Step-by-step explanation:

    x = t + (1/t), y = t² + (1/t²)

    If we square x:

    x² = t² + 2 + (1/t²)

    x² = y + 2

    When x = 3, y = 7.

    Taking derivative with respect to time:

    2x = dy/dx

    dy/dx = 6

    So the equation of the tangent line is:

    y - 7 = 6 (x - 3)

    y - 7 = 6x - 18

    y = 6x - 11
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