Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 4 + ln (t), y = t^2 + 6, (4, 7)

y = 2x − 1

Step-by-step explanation:

By eliminating the parameter, first solve for t:

x = 4 + ln(t)

x − 4 = ln(t)

e^(x − 4) = t

Substitute:

y = t² + 6

y = (e^(x − 4))² + 6

y = e^(2x − 8) + 6

Taking derivative using chain rule:

dy/dx = e^(2x − 8) (2)

dy/dx = 2 e^(2x − 8)

Evaluating at x = 4:

dy/dx = 2 e^(8 − 8)

dy/dx = 2

Writing equation of line using point-slope form:

y − 7 = 2 (x − 4)

y = 2x − 1

Now, without eliminating the parameter, take derivative with respect to t:

x = 4 + ln(t)

dx/dt = 1/t

y = t² + 6

dy/dt = 2t

Finding dy/dx:

dy/dx = (dy/dt) / (dx/dt)

dy/dx = (2t) / (1/t)

dy/dx = 2t²

At the point (4, 7), t = 1.  Evaluating the derivative:

dy/dx = 2(1)²

dy/dx = 2

Writing equation of line using point-slope form:

y − 7 = 2 (x − 4)

y = 2x − 1

based on context clues you gave i'm assuming the answer would be c.