How to find all points for which the tangent line of a parametric equation (x, y, z) passes through a pointFor the function x → ( t ) = ( 2 t + 3 2 − t t 3 − 2 t 2 + t ) t ≥ 0Find all points x → ( t 0 ) for which the tangent line passes through the point (1, 3, 0).I understand how I would go about solving the problem if I had a normal polynomial function or only x and y values. However, with x, y, and z values I am not sure how to approach the question.I know that the derivative of the function is x → ′ ( t ) = ( 2 − 1 3 t 2 − 4 t + 1 ) Would I then have to find the equation of the tangent line and use that with the given point to find the other points?

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How to find all points for which the tangent line of a parametric equation (x, y, z) passes through a pointFor the function            x      →        (  t  )  =      (                            2          t          +          3                                      2          −          t                                                  t            3                    −          2                      t            2                    +          t                      )    t  ≥  0Find all points             x      →        (      t    0    ) for which the tangent line passes through the point (1, 3, 0).I understand how I would go about solving the problem if I had a normal polynomial function or only x and y values. However, with x, y, and z values I am not sure how to approach the question.I know that the derivative of the function is                    x        →              ′    (  t  )  =      (                            2                                      −          1                                      3                      t            2                    −          4          t          +          1                      )  Would I then have to find the equation of the tangent line and use that with the given point to find the other points?

Marisa Colon · Accepted Answer

Hint:Just write the vectors             x      →        (  t  )  −  (  1  ,  3  ,  0  ) and                     x        →              ′    (  t  ) are collinear:            2      t      +      3      −      1        2    =            −      t      −      1              −      1        =                    t        3            −      2              t        2            +      t              3              t        2            −      4      t      +      1          ⟺    t  +  1  =                    t        3            −      2              t        2            +      t              3              t        2            −      4      t      +      1        =            t      (      t      −      1              )        2                    (      3      t      −      1      )      (      t      −      1      )        =  ⋯

For the function vec(x) (t)=((2t+3),(2-t),(t^3-2t^2+t)) t >=0 Find all points vec(x) (t_0) for which the tangent line passes through the point (1, 3, 0).

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