Lines tangent to graph of equation So I have the following problem and can't seem to arrive at the official answer:

Consider the curve whose equation is $$(2-x)y^2 = x^3$$
  Obtain the equations of the lines tangent to the graph of the curve at the points where $x=3/2$.

My attempt
My first thought is that you can isolate y in this equation and then have $y$ as a function of $x$ by the function $$f(x) = \sqrt{(x^3/(-x+2)^2)}$$
But when you turn the equation into a function then obviously you can't have a single $x$ with two outputs, therefore you can't have more than one line tangent to the curve at $x=3/2$. I find this confusing. You can have the equation become a function but you can still have the equation with two outputs for one x.
So I decided to just use implicit differentiation here. Then the problem boils down to finding $$y - y(3/2) = y'(3/2)(x - 3/2)$$
So after a fair amount of work you find $y(3/2)$ and $y'(3/2)$. I would share what I got but really I'd like to see if other people come up with other results.
Official answers are $$y = 3\sqrt{3}x - 3\sqrt{3}$$ and $$y = -3\sqrt{3}x + 3\sqrt{3}$$
 A: You are on the right track, consider
\begin{eqnarray}
\frac{{\rm d}}{{\rm d}x}[(2-x)y^2] &=& \frac{{\rm d}}{{\rm d}x}[x^3] \\
(2-x)\frac{{\rm d}y^2}{{\rm d}x} -y^2 &=& 3x^2 \\
2(2-x)y \frac{{\rm d}y}{{\rm d}x} &=& 3 x^2 \tag{1}
\end{eqnarray}
From the original equation, evaluate $x=3/2$ that gives you
$$
(2 - 3/2)y^2 = (3/2)^2 ~~~\Rightarrow ~~~ y^2 = 2 (3/2)^3 = 27/4
$$
That is
$$
y_{\pm} = 3\sqrt{3}/2 \tag{2}
$$
You need to evaluate these two values into equation (1), that is evaluate the pairs $(x,y) = (3/2, 3\sqrt{3}/2)$ and $(x,y) = (3/2, -3\sqrt{3}/2)$. Can you take it from here?
A: Note that fo $x\ne 2$
$$(2-x)y^2 = x^3\implies y^2=\frac{x^3}{2-x}\implies 2ydy=-\frac{2(x-3)x^2}{(2-x)^2}dx \implies\frac{dy}{dx}=-\frac{2(x-3)x^2}{2y(2-x)^2}$$
and for $x=\frac32$
$$y^2=\frac{x^3}{2-x}=\frac{\frac{27}{8}}{\frac{1}{2}}=\frac{27}{4}\implies y={\pm} \frac{3\sqrt{3}}2 $$
and we obtain


*

*$(x,y)=\left(\frac32,\frac{3\sqrt{3}}2\right)\implies \frac{dy}{dx}=3\sqrt 3$

*$(x,y)=\left(\frac32,-\frac{3\sqrt{3}}2\right)\implies \frac{dy}{dx}=-3\sqrt 3$
and finally


*

*$\left(y-\frac{3\sqrt{3}}2\right)=3\sqrt 3\left(x-\frac{3}2\right)\implies y=3\sqrt 3x-\frac{9\sqrt{3}}2+\frac{3\sqrt{3}}2\implies y=3\sqrt 3x-3\sqrt{3}$

*$\left(y+\frac{3\sqrt{3}}2\right)=-3\sqrt 3\left(x-\frac{3}2\right)\implies y=-3\sqrt 3x+\frac{9\sqrt{3}}2-\frac{3\sqrt{3}}2\implies y=-3\sqrt 3x+3\sqrt{3}$
A: If you move all of the terms to one side of the equality, you get a description of the curve as a level curve of a function $f:\mathbb R^2\to\mathbb R$. The gradient of a function is everywhere normal to its level curves, so this suggests using the point-normal form of the equation of the tangent at $(x_0,y_0)$: $$\nabla f(x_0,y_0)\cdot(x-x_0,y-y_0)=0.$$ For this problem, we have $$f(x,y) = x^3+(x-2)y^2 \\ \nabla f = \left(3x^2+y^2,2(x-2)y\right)$$ and so an equation of the tangent at $(x_0,y_0)$ is $$\left(3x_0^2+y_0^2\right)\left( x-x_0\right)+2(x_0-2)y_0(y-y_0) = 0.$$ When $x=\frac32$, $y=\pm\frac32\sqrt3$. Plug these values into the above equation and simplify, and you’ll get exactly the two Official answers.
