Find the radius of curvature at the origin 
Find the radius of curvature at the origin for the curve $3x^2+4y^2=2x$.

Below is my attempt:
We know that the radius of curvature $\rho=\dfrac{{(1+y_1^2)}^{\frac{3}{2}}}{y_2}$ .
Computing $y_1$ at $(0,0)$ : Differentiating we get $3x+4y\frac{dy}{dx}=1\implies \dfrac{dy}{dx}=\dfrac{1-3x}{4y}.$
How can I evaluate $\dfrac{dy}{dx}$ at the point $(0,0)$?
Please help.
 A: Notice that the graph of the equation $$3x^2 -2x +4y^2=0$$ is an ellipse passing through the origin and tangent to the y-axis therefore $dy/dx$ does not exist.      We use implicit differentiation to find $ x'= dx/dy$ and $x''=d^2x/dy^2$at the origin.
 We get $$6xx'-2x'+8y=0$$and $$6(x')^2+6xx''-2x''+8=0$$Now we evaluate $ x'$ and $x''$ at the origin using the above results. We get $x'=0$ and $x''=4.$ Therefore the curvature at the origin is $$\kappa=\frac{|x''|}{(1+x'^{2})^{\frac{3}{2}}} =4.$$ Therefore the radius of curvature at the origin is $$R=1/4$$
A: The curve is an ellipse centered at $\left(\frac13,0\right)$. You can’t evaluate ${dy\over dx}$ at the origin because the tangent there is vertical, so you’ll have to try something else. For instance, you could parameterize the curve as $x(t) = \frac13(1-\cos t)$, $y(t)=\frac1{2\sqrt{3}}\sin t$ and use the formula $\kappa = {|x'y''-y'x''| \over (x'^2+y'^2)^{3/2}}$ for the curvature. We have $$x'(0) = 0, \; y'(0) = \frac1{2\sqrt{3}} \\ x''(t) = \frac13, \; y''(t) = 0 \\ x'y''-y'x'' = -{1\over 6\sqrt3} \\ x'^2+y'^2 = \frac1{12}$$ and so $\kappa = 4$ and the radius of curvature is $\frac14$.  
If you don’t happen to spot the simple parameterization, it’s also possible to compute the curvature from first- and second-order partial derivatives of $F(x,y)=3x^2+4y^2-2x$. If we let $H(F)$ stand for the Hessian matrix of $F$ and $T(F)$ be the unit tangent $(-F_y,F_x) / \, \|\nabla F\|$, then $$\kappa = {|T(F)H(F)T(F)^T| \over \|\nabla F\|}.$$ For this problem, we have $$\nabla F[0,0]=(-2,0) \\ T(F)[0,0] = (0,-1) \\ H(F)[0,0] = \operatorname{diag}(6,8)$$ so $\kappa = 8/2 = 4$ as before.
A: If we consider the differential $(6x-2)dx+8ydy$, then at $(0,0)$, we see $(6x-2) \neq 0$. By the implicit function theorem, we may write $x$ as a function of $y$. 
Complete the square:
$$3x^2 -2x +4y^2=0$$
$$3(x-\frac{1}{3})^2+4y^2=\frac{1}{3}$$
From here we get:
$$x(y)=\frac{1}{3}+\sqrt{\frac{1}{9}-\frac{4}{3}y^2}$$ 
Let $r:(-\epsilon,\epsilon) \rightarrow \mathbb{R}^{2}$ be given by 
$$r(y) =(x(y),y)$$
The curvature is given by the general formula:
$$\kappa=\frac{|x'y''-x''y'|}{(x'^{2}+y'^{2})^{\frac{3}{2}}}$$
Using our parameterization, we have:
$$\kappa(y)=\frac{|\frac{d^2x}{dy^2}|}{(1+(\frac{dx}{dy})^2)^\frac{3}{2}}$$
Now $\frac{dx}{dy}=0$ when $y=0$. If we compute $\frac{d^2
x}{dy^2}$ we have:
$$\frac{d^2
x}{dy^2}=y(junk)+\frac{4}{3}\bigg(\frac{1}{9}-(\frac{4}{3}y^2)\bigg)^\frac{-1}{2}$$
Evaluating at $y=0$ we find,
$$\kappa(0)=4$$.
Thus $\rho= \frac{1}{\kappa}=\frac{1}{4}$
A: Since, $$\frac{dy}{dx}$$is not defined here, i.e., y axis is tangent to the curve.
No one above is talking about Newtonian method, so I am going with that - it says if curve passes through origin and y axis(or x axis) is tangent to the curve, then radius of curvature at origin = $$lim(x \rightarrow 0) \frac{y²}{2x}$$
We have, $$3{x}^2 - 2x + 4{y}^2 = 0 $$
Dividing it by 2x,
$$\frac{3x}{2} - 1 + 4\frac{y^2}{2x} =0$$
When x --> 0,
$$ -1 + 4\rho = 0$$
$$ \rho = \frac{1}{4}$$
