The equation of the tangent to the curve at the point corresponding to the given value of the parameter will be x + y = c.
The curves are given below.
x = t⁵ + 1 and y = t⁶ + t
Then differentiate the functions with respect to t, then we have
[tex]\rm \dfrac{dx}{dt} = 5t^4\\[/tex] ...1
and
[tex]\rm \dfrac{dy}{dt} = 6t^5 + 1[/tex] ...2
Divide equation 2 by equation 1, then we have
[tex]\rm \dfrac{\dfrac{dy}{dt} }{\dfrac{dx}{dt} } = \dfrac{dy}{dx} = \dfrac{6t^5 + 1}{5t^4}[/tex]
Then the slope of the equation at t = −1, then we have
dy/dx = [6(-1)⁵ + 1]/[5(-1)⁴]
dy/dx = (-6+1)/5
dy/dx = -5/5
dy/dx = -1
Then the equation of the tangent line will be
y = -x + c
x + y = c
Where c is a constant.
More about the tangent of the parameter curves at a point link is given below.
https://brainly.com/question/12648555
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