If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then find the largest possible value of $|a-b|$ I came across the following problem that says:  

If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then the largest possible value of $|a-b|$ is which of the following?
  $(1)\sqrt 7$,  $(2)\sqrt{7/2}$ , $(3)\sqrt {35}$  $(4)7$  

My Attempt:  We notice, $(a-b)^2=7-(4ab+b^2)$ and hence $|a-b|=\sqrt {7-(4ab+b^2)}$ and so $|a-b|$ will be maximum whenever $(4ab+b^2)$ will be minimum. But now I am not sure how to progress further hereon.Can someone point me in the right direction? Thanks in advance for your time.

The previous solution was wrong, because I used $\frac {da}{db} = -1$ instead. I've added an explanation of why we should have used $\frac {da}{db} = 1$.

The simplest approach I can think of, is to realize that you have a conic section, which is an ellipse.
Because you are interested in extreme values of $a-b = K$, this would be lines of the form $a = b + K$. From calculus, it follows that the extremum occurs at the points of your ellipse where $1 = \frac {da}{db}$. By implicit differentiation, 
$$ 2a \frac {da}{db} + 2a + 2b \frac {da}{db} + 4b = 0,$$ 
hence $a = -\frac {3b}{2}$. Plugging this back into the equation, we obtain $\frac {9b^2}{4} - 3b^2 + 2b^2 = 7$, or that $b = \pm 2 \sqrt{\frac {7}{5}}$. Check the corresponding values of $a$ are given by $ ( a, b) = ( -3 \sqrt{ \frac {7}{5} }, 2 \sqrt{ \frac {7}{5} }), (  3\sqrt{\frac {7}{5}}  , -2 \sqrt{ \frac {7}{5}})$. Hence, the maximum of $|a-b|$ is $\sqrt{ 35}$.
