skylarhorse380 skylarhorse380
  • 09-07-2021
  • Mathematics
contestada

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

Respuesta :

Cricetus
Cricetus Cricetus
  • 11-07-2021

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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