Problem:If The equation $ax^2+4xy+y^2+ax+3y+2=0$ represents a parabola then find the value of $a$.
My attempt-I known that in a parabola($e=1$)[where $e$ is eccentricity].So the distance of any general point on any Conic from the directrix($ax+by+c=0$) must be equal to its distance from the focus($\alpha,\beta$).That means one can write the general equation of any conic in form of-
$$(x-\alpha)^2+(y-\beta)^2=e^2\frac{(ax+by+c)^2}{(a^2+b^2)}$$
This means that I have to somehow factorise the given equation in above form and then apply the necessary conditions for.But I haven't been able to get anything useful.
Can Anyone Help me a little bit with this? A Hint will also Suffice.
Best Answer
HINT :
Expand and simplify $$(A^2+B^2)((x-\alpha)^2+(y-\beta)^2)=(Ax+By+C)^2$$ to the form $$Dx^2+Exy+Fy^2+Gx+Hy+\color{red}{2}=0$$ Then, solve the following system : $$D=a,\quad E=4,\quad F=1.$$