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.

As the distance between points is invariant 1,2 under Rotation,
using Rotation of axes by putting $a=x\cos\theta-y\sin\theta$ and $b=x\sin\theta+y\cos\theta$ in $a^2+2ab+2b^2=7$
$(x\cos\theta-y\sin\theta)^2+2(x\cos\theta-y\sin\theta)(x\sin\theta+y\cos\theta)+2(x\sin\theta+y\cos\theta)^2=7$
or, $x^2(\cos^2\theta+2\cos\theta\sin\theta+2\sin^2\theta)$
$+xy(-2\cos\theta\sin\theta+2\cos^2\theta-2\sin^2\theta+4\cos\theta\sin\theta)+y^2(\sin^2\theta-2\cos\theta\sin\theta+2\cos^2\theta)=7$
Applying $\cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta$ and $\sin2\theta=2\cos\theta\sin\theta,$
$x^2(3-\cos2\theta+2\sin2\theta)+2xy(2\cos2\theta+\sin2\theta)+y^2(3+\cos2\theta-2\sin2\theta)=14$
To remove $xy$ term we put $2\cos2\theta+\sin2\theta=0\implies \frac{-\sin2\theta}2=\frac{\cos2\theta}1=\pm\frac1{\sqrt{2^2+1^2}}$
Taking the '$+$' sign and on simplification  we get $\frac{x^2}{\frac{7(3+\sqrt5)}2}+\frac{y^2}{\frac{7(3-\sqrt5)}2}=1$
Comparing with $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1,$
the extreme values of $x$ are $\pm\sqrt{\frac{7(3+\sqrt5)}2}=\pm\frac{\sqrt7}2\sqrt{6+2\sqrt5}=\pm\frac{\sqrt7}2\sqrt{(\sqrt5+1)^2}=\pm\frac{\sqrt{35}+\sqrt7}2$
Similarly, the respective extreme values of $y$ are $\pm\frac{\sqrt{35}-\sqrt7}2$
So, maximum value of $|a-b|=\left|\frac{\sqrt{35}+\sqrt7}2-(\frac{\sqrt7-\sqrt{35}}2)\right|=\sqrt{35}$
