Lines Do Intersect, You Know!

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 patial differentiating the equation wrt x and wrt y, and then solving the lines obtained.

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

So $\frac{af-gh}{h^2-ab}=0$

$af=gh$

$\large{\to 6\times \frac{3}{2}=-12\times \frac{-\alpha}{2}}$

$\large{\boxed{\alpha=\frac{3}{2}}}$

Substituting $y=0$ in the given equation

$6x^2-24x+\beta=0$

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

$D=0$

$576-4\times 6 \times \beta=0$

$\large{\boxed{\beta=24}}$

If you don't remember or don't know how to derive the equations for the product of two lines, this can be solved with factorization.

Since the equation represents the product of two lines, use the Cartesian line representation.

Let $L_1: ax + by + c = 0$ .

In order to match the $x^2, y^2, \beta$ terms of the given equation, $L_2: \frac{6}{a}x + \frac{-3}{b}y + \frac{\beta}{c} = 0$

$6x^2 - \alpha xy - 3y^2 - 24x + 3y + \beta = 0 \\ (ax + by + c)(\frac{6}{a}x + \frac{-3}{b}y + \frac{\beta}{c}) = 0 \\ 6x^2 + (\frac{-3a}{b}+\frac{6b}{a}) xy - 3y^2 + (\frac{a \beta}{c}+\frac{6c}{a})x + (\frac{b \beta}{c}-\frac{3c}{b})y + \beta = 0$ [1]

Since $L_1$ and $L_2$ intersect on the x-axis, set y = 0 and find x

$ax + c = \frac{6}{a}x + \frac{\beta}{c} = 0 \Rightarrow x = \frac{-c}{a} = \frac{-a \beta}{6c} \Rightarrow \frac{6c}{a} = \frac{a \beta}{c}$ [2]

Substitute [2] into the x term of [1] to get $\frac{a \beta}{c}+\frac{6c}{a} = \frac{6c}{a}+\frac{6c}{a} = \frac{12c}{a} = -24 \Rightarrow \underline{c = -2a}$ [3]

Substitute [3] into the x term of [1] to get $\frac{a \beta}{c}+\frac{6c}{a} = \frac{a \beta}{-2a}+\frac{-12a}{a} -24 \Rightarrow \frac{\beta}{2} + 12 = 24 \Rightarrow \underline{\beta = 12}$ [4]

From [1] we have the following: $\frac{-3a}{b}+\frac{6b}{a} = -\alpha, \frac{b \beta}{c}-\frac{3c}{b} = 3$ [5]

Substitute $\beta, c$ into [5] to get

$\frac{24b}{-2a}+\frac{6a}{b} = 3 \\ \frac{-3a}{b} + \frac{6b}{a} = \frac{-3}{2}\\ -\alpha = \frac{-3}{2} \Rightarrow \underline{\alpha = \frac{3}{2}}$

$\therefore \boxed{20 \alpha - \beta = 20(\frac{3}{2}) - 24 = 6}$

As is given, the lines intersect on the x-axis, let the point of intersection of the lines be $\displaystyle P(k,0)$ .

Therefore, point $P$ must satisfy the given equation:

$\displaystyle \Rightarrow 6k^2 - 24k +\beta =0$

There is only one such $k$ possible for the intersection of the two lines and so the above equation must have repeated roots:

$\displaystyle \Rightarrow b^2 - 4ac = 0$

$\displaystyle \Rightarrow \boxed{\beta=24}$ and $\displaystyle \boxed{k=2}$

Substitute $\displaystyle (2,y)$ back into the equation:

$\displaystyle \Rightarrow 6 (2)^2 - \alpha(2)y + y^2 + 3y - 24(2) + 24 = 0$

$\displaystyle \Rightarrow y^2 + (3 - 2\alpha)y = 0$

But we know that $\displaystyle y=0$ is the only solution to the above equation, so even this equation has repeated roots:

$\displaystyle \Rightarrow b^2 - 4ac = 0$

$\displaystyle \Rightarrow \boxed{\alpha=\frac{3}{2}}$

$\displaystyle \Rightarrow 20\alpha - \beta = 20(\frac{3}{2}) - 24 = \boxed{6 }$