Finding Tangent Line Eq.

Question

Find the equation of the tangent line to the parametric curve given by x= 1/( t+1) , y= t / (t−1) for t = 2.

Answer 1

When t = 2, x = 1/3, y = 2.

dy/dx=(dy/dt)/(dx/dt)= (-1/(t-1)²)/(-1/(t+1)²)= (t+1)²/(t-1)²

Evaluating dy/dx when t = 2:

3²/1²=9.

So the equation of the tangent line at the point where t = 2:

y= 9(x-(1/3)) + 2

y = 9x-1

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Answer 2

In this kind of problem, aside from the point (x₁,y₁), we also have to solve for the slope of the tangent line using derivative:

x= 1/( t+1) , y= t / (t−1) for t = 2.

x = 1/(2+1) = 1/3

y = 2/(2-1) = 2

Therefore (x₁,y₁) = (1/3,2).

Now let's find the dy/dx for the slope. First:

t + 1 = 1/x

t = 1/x -1

substitute t in the equation with y:

y = (1/x - 1)/(1/x -1 -1) = (1/x - 1)/(1/x -2)

y = [(1-x)/x] / [(1-2x)/x]

y = (1-x)/(1-2x)

Then differentiate:

dy/dx = [(1-2x)(-1) - (1-x)(-2)] / (1-2x)²

dy/dx = 1 / (1-2x)²

When x = 1/3, dy/dx = 9

Using the point-slope form of linear equation:

y - y₁ = m (x - x₁)

y - 2 = 9 (x - 1/3)