Range of quadratic function using discriminant

Question

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?

Answer 1

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 $\dfrac xy\le-2+1$

Answer 2

Consider the curves $x^2-2xy-3y^2=4 \ (C)$ and $2x^2-2xy+y^2=k^2+4 \ (E)$ At the points where they intersect, we can substitute the value of $-2xy$ from the first into the second to get $$x^2+4y^2=k^2 \ (E)$$ This represents an ellipse centered at the origin. Now, either with a bit of graph sketching, or by considering $(C)$ as a quadratic in $x^2$, deduce that $y$ can range anywhere in the real numbers, and that $(C)$ is a hyperbola. Increasing $k$ in $(E)$ just enlarges the ellipse, and it’s not hard to see that there is no upper bound on $k$ for $(E)$ and $(C)$ to intersect. For obtaining a lower bound, we need to find $k$ such that the two curves touch. Note that if $(x,y)$ lies on the two curves, then so does $(-x,-y)$. So we’re looking for exactly two intersections. Then substituting $x=\pm\sqrt{k^2-4y^2}$ in $(C)$ gives a quadratic in $y^2$: $$65y^4-y^2(18k^2-56)+(k^2-4)^2 = 0 $$ Setting the discriminant equal to zero will give $k^2=\frac{\sqrt{65}-1}{2}$ and hence we have the desired range: $$\frac{\sqrt{65}+7}{2} \le 2x^2-2xy+y^2 =k^2+4\lt \infty $$

What was wrong in your approach? What you stated was a necessary condition on $a$, but that doesn’t mean that $a$ could really equal anything in that range.

Answer 3

Others have pointed out where you went wrong. I just want to provide an alternative proof.

$$x^2-2xy-3y^2=4=(x+y)(x-3y)$$

Denote $z=x+y, w=x-3y$, then $zw=4, x=(3z+w)/4, y=(z-w)/4$.

We want to find the range of $$2x^2-2xy+y^2=\frac{1}{16} (5w^2+14wz+13z^2)=\frac 72+\frac{1}{16} (5w^2+13z^2) \tag 1$$ under the constraint $zw=4$.

Clearly $(1)$ can be as large as you want. And if you apply AM-GM you get the minimum $$\frac 72+\frac{1}{16} (5w^2+13z^2) \ge \frac 72 + \frac{1}{16} 2 \sqrt{5} \sqrt{13}|wz|=\frac{7+\sqrt{65}}{2}$$