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


1 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


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 $t=\pi/4$

$$...$$ enter image description here

  • $\begingroup$ 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/… $\endgroup$ Jan 31, 2021 at 20:28
  • $\begingroup$ @user9106985 $y-\frac{1}{\sqrt{2}}+1=\frac{x-\sqrt{2}}{\sqrt{2}-1}$ is not tangent to the curve $\endgroup$
    – Raffaele
    Jan 31, 2021 at 20:46
  • $\begingroup$ 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? $\endgroup$ Jan 31, 2021 at 20:50
  • $\begingroup$ And did you reproduce that plot in wolfram? Thanks in advance. $\endgroup$ Jan 31, 2021 at 20:51
  • 1
    $\begingroup$ @user9106985 $$y+3=\left(\sqrt{2}+1\right) x$$ it is even simpler in this form $\endgroup$
    – Raffaele
    Jan 31, 2021 at 20:58

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