<s>[INST] Range of quadratic function using discriminant 
Let $x^2-2xy-3y^2=4$. Then find the range of $2x^2-2xy+y^2$.

Let $2x^2-2xy+y^2=a$.
Then $ax^2-2axy-3ay^2=4a=8x^2-8xy+4y^2\implies (a-8)x^2-(2a-8)xy-(3a+4)y^2=0$.
We divide both side by $y^2$ and let $t=\frac{x}{y}$.
Then it implies $(a-8)t^2-(2a-8)t-(3a+4)=0$.
Since its discriminant is not negative, $\frac{\Delta}{4}\ge 0\implies a^2-7a-4\ge 0$. It gives us $a$ can have negative values like $-1$. But if clearly contracts $a-4=x^2+4y^2\ge 0\implies a\ge 4$. Where did I mistake?
 [/INST] Method$\#1:$
For real $t,$ $$a^2-7a-4\ge0$$
If  $x=ty$
$$4=x^2-2xy-3y^2=y^2(t^2-2t-3)$$
$$t^2-2t-3=\dfrac4{y^2}>0$$
$$\implies(t-3)(t+1)>0$$
Either $t<-1$ or $t>3$
So, the values of $a$ must satisfy this condition as well
Method$\#2:$
$$4=(x-y)^2-(2y)^2$$
WLOG $y=\tan t, x-y=2\sec t\implies x=2\sec t+\tan t$
$$a=2(2\sec t+\tan t)^2-2(2\sec t+\tan t)\tan t+\tan^2t$$
Multiplying both sides  by $\cos^2t=1-\sin^2t,$
$$(1+a)\sin^2t+4\sin t+8-a=0$$
What if $a+1=0?$
Else
$$\sin t=\dfrac{-2\pm\sqrt{a^2-7a-4}}{a+1}$$
As $\sin t$ is real, the discriminant must be $\ge0$
But that is not sufficient, we need $$-1\le\dfrac{-2\pm\sqrt{a^2-7a-4}}{a+1}\le1$$ as well for real $t$
Also, $\dfrac xy=2\csc t+1$ $\implies\dfrac xy\ge2+1$ or $\dfrac xy\le-2+1$
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