Find the equations of the tangents to the curve x=9t2+3, y=6t3+3 that pass through the point (12,9). July 22, 2021 by Thành Đạt Find the equations of the tangents to the curve x=9t2+3, y=6t3+3 that pass through the point (12,9).
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] Reply
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]