<s>[INST] 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.
 [/INST] 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 $|a-b|=\left|\frac{\sqrt{35}+\sqrt7}2-(\frac{\sqrt7-\sqrt{35}}2)\right|=\sqrt{35}$
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