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I'm asked to find the equation of the normal line for this implicitly given curve

Find an equation of the normal line to the curve $$ \ln(2-y)=\arctan(xy)-\tan(3x)$$ at the point where $x=0$.

I can do the question fine if it was the equation of a tangent line, but how would I do the normal line?

All I can really do is given that $x=0$, I can get that $y=1$. I can then get the slope of the tangent line at $(0,1)$ which is $y'=2$. Then because the slope of the normal line would be the negative reciprocal, I get that $y'_\perp=\frac{-1}{2}$.

However, to find the equation of the normal line, I need a point on that normal line, and I have no idea how to get that. Any advice or suggestion would be useful!

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  • $\begingroup$ You know that the normal passes through x = 0. Substitute the value of x in the original equation and get the value of y. Use these to frame the tangent equation. $\endgroup$
    – Sam
    Commented Oct 2, 2019 at 7:27
  • $\begingroup$ I might miss understood your question, but , shouldn't the normal line pass through $(0,1)$? $\endgroup$
    – Chaos
    Commented Oct 2, 2019 at 7:28

1 Answer 1

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The normal line when $x=0$ is nothing but the line with slope $-\frac 1 2$ passing through $(0,1)$ which is $y=-\frac 1 2 x+1$.

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