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

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

step-by-step explanation: