Find intersection point of pair of lines $a(x-\alpha)^2+h(x-\alpha)(y-\beta)+b(y-\beta)^2$

Question

The general equation for a pair of lines is $$ax^2+2hxy+by^2=0$$ if they intersect at (0,0). Clearly, the origin has been shifted to some other point. We write it $x=X+h,y=Y+k$ if the origin is shifted to (h,k). Then in the given equation, the x and y terms are in the form of $X+(-\alpha),Y+(-\beta)$. Hence the origin is shifted to $(-\alpha,-\beta)$

But the answer given is $(\alpha,\beta)$

Answer 1

Let $u=x-\alpha$ and $v=y-\beta$. Then, the given equation becomes

$$au^2+huv+bv^2=0$$

Since the point of intersection for this pair of lines is known to be $(u,v)=(0,0)$, the corresponding $xy$-point of intersection is $(x,y)=(u+\alpha,v+\beta)= (\alpha,\beta)$.