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I have not seen a problem like this so I have no idea what to do.

Find an equation of the tangent to the curve at the given point by two methods, without elimiating parameter and with.

$$x = 1 + \ln t,\;\; y = t^2 + 2;\;\; (1, 3)$$

I know that $$\dfrac{dy}{dx} = \dfrac{\; 2t\; }{\dfrac{1}{t}}$$

But this give a very wrong answer. I am not sure what a parameter is or how to eliminate it.

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