# Parametric Tangent Problem driving me insane

## Homework Statement

$$x = e^{t} , y = (t-1)^{2} , (1,1)$$
Find an equation of the tangent to the curve at a given point by two methods. Without eliminating the parameter and by first eliminating the parameter.

The answer in the book says $$y = -2x + 3$$ and I cannot see how you get it.

So I can't do it either way.

## Homework Equations

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

## The Attempt at a Solution

$$\frac{dy}{dt} = 2(t-1)$$

$$\frac{dx}{dt} = e^{t}$$

$$\frac{dy}{dx} = \frac{2(t-1)}{e^{t}}$$

okay so now I need to substitute in (1,1)

Rearranging x and y in terms of t does no good:

$$x = e^{t}$$

so$$t = lnx ...(1)$$

$$\sqrt{y} = t - 1$$

so $$t = \sqrt{y} + 1 ... (2)$$

At point (1,1):

t = 0 by equation (1)

t = 2 by equation (2)

So which t to use?

Lets try a different approach:

$$\frac{dy}{dx} = \frac{2(\sqrt{y} + 1 - 1)}{e^{lnx}}$$

so $$\frac{dy}{dx} = \frac{2\sqrt{y}}{x}$$

Substituting (1,1)

$$\frac{dy}{dx} = 2 = m$$

so $$y - y_{1} = m(x - x_{1})$$

$$y = 2x - 1$$

I really can't see what I'm doing wrong. Any help appreciated.

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Ben Niehoff
Gold Member
$$t = lnx ...(1)$$

$$\sqrt{y} = t - 1$$

so $$t = \sqrt{y} + 1 ... (2)$$

At point (1,1):

t = 0 by equation (1)

t = 2 by equation (2)

So which t to use?
Remember every positive number has two real square roots. Try using the other one.

Ah of course. That sorts it out:

$$\frac{dy}{dx} = \frac{2(0-1)}{e^{0}}$$

$$\frac{dy}{dx} = -2$$

$$y - 1 = -2x + 2$$
$$y = -2x + 3$$

Thanks!