# Equation of a line normal to an implicit function

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!

• 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.
– Sam
Commented Oct 2, 2019 at 7:27
• I might miss understood your question, but , shouldn't the normal line pass through $(0,1)$? Commented Oct 2, 2019 at 7:28

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