Find the implicit function for $xy^2+4x^2y-12=0$ at $(1,2)$. Then find the tangent and the normal to $y=\varphi(x)$ at $x=1$.
Let $y=\varphi(x)$ (1)
Now: $x\varphi(x)^2+4x^2\varphi(x)-12=0$ $\Rightarrow \frac{d}{dx} = \varphi^2(x)+2x\varphi'(x)\varphi(x)+8x\varphi(x)+4x^2\varphi'(x)=0$
Using again (1) $\Rightarrow 4+4x\varphi'(x)+16x+4x^2\varphi'(x)=0$
$\Rightarrow \varphi'(x)=\frac{-4x-1}{x+x^2}$
Now, I don't know how to find neither the tangent nor the normal. How'd I find them?
I am assuming the problem wants the tangent line to the graph of $xy^{2} + 2x^{2}y - 12 = 0$. As you calculated, $\frac{d}{dx} (xy^{2} + 2x^{2}y - 12) = y^{2} + 2xy y' + 4x^{2} y' + 8xy = 0 \implies y' = \frac{-8xy-y^{2}}{2xy+4x^{2}}$. To find the tangent line at $(1,2)$, we need to find the slope, which is given by $y'(1,2) = -\frac{5}{2}$. Knowing that the tangent line has a point of $(1,2)$, we have that using point slope form $(y-2) = -\frac{5}{2}(x - 1) \implies y = -\frac{5}{2}x + \frac{9}{2}$. For the normal line, it is perpendicular to the tangent line and intersects it at $(1,2)$. Performing a similar calculation yields that $y = \frac{2}{5} x +\frac{8}{5}$. Hence the lines are:
$$ y_{T} = -\frac{5}{2}x + \frac{9}{2} \\ y_{N} = \frac{2}{5} x +\frac{8}{5} $$ A graphic of this problem can be found below:
Which curve is the tangent line, normal line and the given function?