Lines do intersect , you know Part 2

Let the two straight lines be: $\space \begin{cases} y = m_1x + c_1 = 0 \\ y = m_2x + c_2 = 0 \end{cases}$

Since they intersect at the $y$ -axis: $\space \Rightarrow c_1=c_2 = c$ . Therefore, we have:

$10x^2- 2\alpha xy + y^2 + 10x + 2y - \beta \\ = (m_1x -y + c)(m_2x - y + c) \\ = m_1m_2x^2 + (m_1+m_2)xy + y^2 + c(m_1+m_2)x + 2cy + c^2$

Equating the coefficients on both sides:

$\begin{cases} m_1m_2 = 10 & ...(1) \\ m_1+m_2 = - 2\alpha & ...(2) \\ c(m_1+m_2) = 10 & ...(3) \\ 2c = 2 & ...(4) \\ c^2 = - \beta & ...(5) \end{cases}$

$\begin{cases} (4): & c = \dfrac{2}{2} = 1 \\ (5): & \beta = - c^2 = - 1 \\ (3): & (m_1+m_2) = \dfrac{10}{1} = 10 \\ (2): & \alpha = - \dfrac {m_1+m_2}{2} = - 5 \end{cases}$

$\Rightarrow |\alpha| + |\beta| = |-5| + |-1| = \boxed{6}$

Point of intersection of two lines represented by

$ax^2+2hxy+by^2+2gx+2fy+c=0$ is

$\large{\left( \frac{bg-hf}{h^2-ab},\frac{af-gh}{h^2-ab} \right)}$

U can easily prove it by partial differentiating the equation wrt x and wrt y, and then solving the equations obtained.

Since they meet at y-axis . Let the point be $(0,t)$

So $\frac{bg-hf}{h^2-ab}=0$

$bg=hf$

$\large{\to 1\times 5=-\alpha \times 1}$

$\large{\boxed{\alpha=-5}}$

Substituting $x=0$ in the given equation

$y^2+2y-\beta=0$

Since the lines intersect at a common point , so the above quadratic equation $\text{must have equal roots}$ .

$D=0$

$4-4\times (-\beta)=0$

$\large{\boxed{\beta=-1}}$

Since they intersect at y axis while putting x=0 remaining quadratic in y must have equal roots. Hence beta=-1 Now applying condition for a second degree equation to be a pair of straight lines i.e. abc+2fgh-af^2-bg^2-ch^2=0 We get quadratic in alpha . Solving we get alpha=-5