How to convert parametric form of tangent line to Cartesian equivalent

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 $$y-y_{0}=f'(x_{0})(x-x_{0})$$?

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 $$t=\pi/4$$

$$...$$

• Thanks for the clarity. Somehow your answer is different to that given in the book. I have checked with wolfram wolframalpha.com/input/… and wolframalpha.com/input/… Jan 31, 2021 at 20:28
• @user9106985 $y-\frac{1}{\sqrt{2}}+1=\frac{x-\sqrt{2}}{\sqrt{2}-1}$ is not tangent to the curve Jan 31, 2021 at 20:46
• Ok... the 2nd link I gave you is the answer I see in the book...so I guess it is a typo in the book. How would you explicitly write your answer in the Cartesian form as in my post? Jan 31, 2021 at 20:50
• And did you reproduce that plot in wolfram? Thanks in advance. Jan 31, 2021 at 20:51
• @user9106985 $$y+3=\left(\sqrt{2}+1\right) x$$ it is even simpler in this form Jan 31, 2021 at 20:58