Finding minimum value of $ \frac{x^2 +y^2}{y} $ 
Finding the minimum value of $\displaystyle \frac{x^2 +y^2}{y}.$ where $x,y$ are 
  real
   numbers satisfying $7x^2 + 3xy + 3y^2 = 1$

Try: Equation $7x^2+3xy+3y^2=1$ represent Ellipse
with center is at origin.
So substitute $x=r\cos \alpha $ and $y=r\sin \alpha$ 
in $7x^2+3xy+3y^2=1$
$$3r^2+4r^2\cos^2 \alpha+3r^2\sin \alpha \cos \alpha =1$$
$$3r^2+2r^2(1+\cos 2 \alpha)+\frac{3r^2}{2}\sin 2 \alpha =1$$
$$8r^2+r^2(4\cos 2 \alpha+3\sin \alpha)=2$$
So $$r^2=\frac{2}{8+(4\cos 2 \alpha+3\sin \alpha)}$$
$$\frac{2}{8+5}=\frac{2}{13}\leq r^2\leq \frac{2}{8-5}=\frac{2}{3}$$
we have to find minimum of $$\frac{x^2+y^2}{y}=\frac{r}{\sin \alpha}$$
How can i find it, could some help me 

Making $y = \lambda x$ we have
$$
\min f(x,\lambda) \ \ \mbox{s. t. }\ \ g(x,\lambda) = 0
$$
here 
$$
\begin{cases}
f(x,\lambda) = \frac{1+\lambda^2}{\lambda}x\\
g(x,\lambda) = x^2(7+3\lambda+3\lambda^2)-1=0
\end{cases}
$$
this minimization problem is equivalent to
$$
\min F(\lambda) = \left(\frac{1+\lambda^2}{\lambda}\right)^2\frac{1}{7+3\lambda+3\lambda^2}
$$
and then
$$
F'(\lambda)= 0\to (1 + \lambda^2) (3 \lambda^3 + 2 \lambda^2- 9 \lambda  -14 )=(\lambda-2) (7 + 8 \lambda + 3 \lambda^2) = 0
$$
hence $\lambda = 2\to x = \pm \sqrt{\frac{1}{7+3\times 2+3\times 2^2}} = \pm\frac{1}{5}$ etc.
