How to convert parametric form of tangent line to Cartesian equivalent

Question

I have calculated the tangent line of the following curve $c(t)=(2cost−cos2t,2sint−sin2t$) @ $t=\pi/4$ to be

$l(t)=(\sqrt{2}, \sqrt{2}-1)+ (t- \frac{\pi}{4})(2-\sqrt{2}, \sqrt{2})$

How do I convert this to $y-y_{0}=f'(x_{0})(x-x_{0})$?

Answer 1

From $$(\sqrt{2}, \sqrt{2}-1)+ \left(t- \frac{\pi}{4}\right)(2-\sqrt{2}, \sqrt{2})$$ we get $$\begin{cases} x=\left(2-\sqrt{2}\right) \left(t-\frac{\pi }{4}\right)+\sqrt{2}\\ y=\sqrt{2} \left(t-\frac{\pi }{4}\right)+\sqrt{2}-1\\ \end{cases} $$ from the second equation we have

$t-\frac{\pi}{4}=\frac{y}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1$

Plug in the first, expand and simplify $$\sqrt{2} x+\left(\sqrt{2}-2\right) y=6-3 \sqrt{2}$$

in another form $$y+3=\left(\sqrt{2}+1\right) x$$ edit

look at the image below. The curve and the tangent at $t=\pi/4$

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