Find an equation of the tangent to the curve at the given point by two methods: (a) without eliminating the parameter and
(b) by first eliminating the parameter.
x = 1 + ln t, y = t2 + 2; (1, 3)

1) y = 2x + 1

2) y = 2x + 1

Step-by-step explanation:

The parametric equation given is;

x = 1 + ln t and y = t² + 2 at (1, 3)

1) without eliminating the parameter;

Using, x = 1 + ln t ;

dx/dt = 1/t

Using y = t² + 2;

dy/dt = 2t

Slope which is dy/dx is gotten from;

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

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

dy/dx = 2t²

For x = 1 + In t, at x = 1, we have;

1 = 1 + In t

In t = 0

t = 1

For y = t² + 2, at y = 3, we have;

3 = t² + 2

t² = 3 - 2

t² = 1

t = ±1

Since t = ±1, then;

dy/dx = 2(±1)²

dy/dx = 2

Equation of the tangent is;

y - 3 = 2(x - 1)

y - 3 = 2x - 2

y = 2x - 2 + 3

y = 2x + 1

2) By eliminating the parameter

x = 1 + In t

Let's make t the subject of the equation.

In t = x - 1

t = e^(x - 1)

Let's put e^(x - 1) for t in y = t² + 2

Thus;

y = e^(x - 1)² + 2

y = e^(2(x - 1)) + 2

Thus, parameter has been eliminated

Equation of the tangent is gotten from;

y - y1 = m(x - x1)

m is gradient = dy/dx = 2e^(2(x - 1))

at (1, 3), we have x = 1. Thus;

m = 2e^(2(1 - 1))

m = 2e^0

m = 2

Thus, equation of tangent at (1,3) is;

y - 3 = 2(x - 1)

y - 3 = 2x - 2

y = 2x - 2 + 3

y = 2x + 1

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