Find the equations of the tangents to the curve x=9t2+3, y=6t3+3 that pass through the point (12,9).

Find the equations of the tangents to the curve x=9t2+3, y=6t3+3 that pass through the point (12,9).

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  1. Answer:

    The equation will be “[tex]y=x-3[/tex]”.

    Step-by-step explanation:

    Given:

    Points (12, 9) = (x, y)

    ⇒ [tex]x=9t^2+3[/tex]

    then,

     [tex]\frac{dy}{dt}=18t[/tex]

    or,

    ⇒ [tex]y=6t^3+3[/tex]

    then,

     [tex]\frac{dy}{dt}=18t^2[/tex]

    ⇒ [tex]\frac{dy}{dx}=\frac{18t^2}{18t}[/tex]

            [tex]=t[/tex]

    By using the point slope form.

    The equation of tangent will be:

    ⇒ [tex]y-9=1(x-12)[/tex]

       [tex]y-9=x-12[/tex]

              [tex]y=x-12+9[/tex]

              [tex]y=x-3[/tex]

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