For $\alpha,\beta,\gamma\in\mathbb{R}$ with $0<\alpha<\beta$, if $$\alpha x^2+4\gamma xy+\beta y^2+4p(x+y+1)=0$$ represent a pair of lines. Then which one is right?
(a) $p\in[\alpha,\beta]$
(b) $p\leq \alpha$
(c) $p\geq \alpha$
(d) $p\in(-\infty,\alpha]\cup [\beta,\infty)$
Try: Camparing the equation
$\alpha x^2+4\gamma xy+\beta y^2+4px+4py+4=0$ with general equation of conic
$ax^2+2hxy+by^2+2gx+2fy+c=0$ we have
$a=\alpha,h=2\gamma,b=\beta, g=2p,f=2p,c=4p$
Now if conic represent pair of lines, Then $h^2-ab=0$
So $4\gamma^2-\alpha \cdot \beta=0$
Now How can i relate $p$ with $\alpha$ and $\beta$.
I am struck at that point.
could some help me , Thanks
This answer shows that there are no correct options and that the range of $p$ is $$p\in\bigg(-\infty,0\bigg]\cup \bigg[\beta,\infty\bigg)$$
For the condition "a pair of lines", we have two cases to consider :
intersecting lines
parallel lines (including "coincident lines")
Here, let
$$\begin{align}\Delta&:=\begin{vmatrix} \alpha & 2\gamma & 2p \\ 2\gamma & \beta & 2p \\ 2p & 2p & 4p \\ \end{vmatrix}=-4p((\alpha+\beta-4\gamma)p+4\gamma^2-\alpha\beta) \\\\J&:=\begin{vmatrix} \alpha & 2\gamma \\ 2\gamma & \beta \\ \end{vmatrix}=\alpha\beta-4\gamma^2 \\\\K&:=\begin{vmatrix} \alpha & 2p \\ 2p & 4p \\ \end{vmatrix}+\begin{vmatrix} \beta & 2p \\ 2p & 4p \\ \end{vmatrix}=4p(\alpha+\beta-2p) \end{align}$$
Let us consider a necessary and sufficient condition in each case (see here for the details) :
The equation represents intersecting lines $$\begin{align}&\iff \Delta=0,J\lt 0 \\\\&\iff p((\alpha+\beta-4\gamma)p+4\gamma^2-\alpha\beta) =0,\ \alpha\beta-4\gamma^2\lt 0 \\\\&\iff \begin{cases}p=0,\ \alpha\beta-4\gamma^2 \lt 0\\\\\quad\text{or}\\\\ p=\frac{4\gamma^2-\alpha\beta}{4\gamma-\alpha-\beta},\ \alpha\beta-4\gamma^2\lt 0\end{cases}\end{align}$$
The equation represents parallel lines (including "coincident lines") $$\begin{align}&\iff \Delta=0,\ J=0,\ K\le 0 \\\\&\iff p((\alpha+\beta-4\gamma)p+4\gamma^2-\alpha\beta) =0,\ \alpha\beta-4\gamma^2=0,\ p(\alpha+\beta-2p) \le 0 \\\\&\iff \begin{cases}p=0,\ \alpha\beta-4\gamma^2=0 \\\\\quad\text{or}\\\\ \alpha+\beta-4\gamma=0,\ \alpha\beta-4\gamma^2=0,\ 0\lt p\le\frac{\alpha+\beta}{2}\end{cases} \\\\&\iff p=0,\ \alpha\beta-4\gamma^2=0 \end{align}$$ (the latter case doesn't happen since then we get $\alpha=\beta$ which contradicts $\alpha\lt \beta$.)
Now, suppose that $p=\frac{4\gamma^2-\alpha\beta}{4\gamma-\alpha-\beta}=\alpha$. This implies $(\alpha-2\gamma)^2=0$ from which $\gamma=\frac{\alpha}{2}$ follows. From $\alpha\beta-4\gamma^2\lt 0$, we get $\alpha\beta-4(\frac{\alpha}{2})^2\lt 0$ implying $\beta\lt\alpha$ which contradicts $\beta\gt\alpha$.
So, we see that $\alpha$ is not included in the range of $p$.
Since every option includes $\alpha$, there are no correct options.
In the following, let us find the range of $p$.
Let us consider the case where $p=\frac{4\gamma^2-\alpha\beta}{4\gamma-\alpha-\beta}$ and $\alpha\beta-4\gamma^2\lt 0$.
If $4\gamma-\alpha-\beta\lt 0$, then $p\lt 0$.
If $4\gamma-\alpha-\beta\gt 0$, then $p\gt 0$, and we have $$p\ge \beta\iff \frac{4\gamma^2-\alpha\beta}{4\gamma-\alpha-\beta}\ge\beta\iff (\beta-2\gamma)^2\ge 0\quad\text{which indeed holds}$$
Finally, let us show that there always exists $(\alpha,\beta,\gamma)$ such that $$\small p\in\bigg(-\infty,0\bigg)\cup \bigg[\beta,\infty\bigg)\quad\text{and}\quad p=\frac{4\gamma^2-\alpha\beta}{4\gamma-\alpha-\beta}\quad\text{and}\quad \gamma\in\left(-\infty,-\frac{\sqrt{\alpha\beta}}{2}\right)\cup\left(\frac{\sqrt{\alpha\beta}}{2},\infty\right)$$
Let $f(x)=\frac{4x^2-\alpha\beta}{4x-\alpha-\beta}$. Then, we have $$f'(x)=\frac{4(2x-\alpha)(2x-\beta)}{(4x-\alpha-\beta)^2},\qquad f\left(\frac{\beta}{2}\right)=\beta$$ with $$\small f\left(\pm\frac{\sqrt{\alpha\beta}}{\sqrt 2}\right)=0,\quad f'\left(\frac{\alpha}{2}\right)=f'\left(\frac{\beta}{2}\right)=0,\quad -\frac{\alpha\beta}{2}\lt\frac{\alpha}{2}\lt\frac{\sqrt{\alpha\beta}}{2}\lt\frac{\alpha+\beta}{4}\lt\frac{\beta}{2}$$
Considering the graph of $y=f(x)$ and noting that $0$ is included in the range of $p$, we see that the range of $p$ is $$\color{red}{p\in\bigg(-\infty,0\bigg]\cup \bigg[\beta,\infty\bigg)}$$