Find the equation of two ... 
Find the single equation of two straight lines that pass through the point $(2,3)$ and parallel to the line $x^2 - 6xy + 8y^2 = 0$.

My Attempt:
Let, $a_1x+b_1y=0$ and $a_2x+b_2y=0$ be the two lines represented by $x^2-6xy+8y^2=0$.
then,
$$(a_1x+b_1y)(a_2x+b_2y)=0$$
$$(a_1a_2)x^2+(a_1b_2+b_1a_2)xy+(b_1b_2)y^2=0$$
Comparing with $x^2-6xy+8y^2=0$,
$a_1a_2=1, -(a_1b_2+b_1a_2)=6, b_1b_2=8$.
I got stuck at here. Please help me to continue this.

The lines $$(x-2)^2 - 6(x-2)(y-3) + 8(y-3)^2 = 0$$ on transfer of origin to $(2,3)$, using the transformations $X = x-2, Y= y-2$ becomes $$X^2 - 6XY + 8Y^2 = 0$$ Hence the required equation is $$(x-2)^2 - 6(x-2)(y-3) + 8(y-3)^2 = 0$$
