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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.

Freelancer
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  • See https://en.wikipedia.org/wiki/Rotation_of_axes#Elimination_of_the_xy_term_by_the_rotation_formula – lab bhattacharjee Mar 21 '16 at 12:51
  • There is an interesting fact: if $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is any nondegenerate conic, then it is a parabola if and only if $B^2-4AC=0$. This shows that your value $a$ must be equal to $4$. – Crostul Mar 21 '16 at 12:55
  • @Crostul I was reading the same thing which you mentioned here from the Wikipedia link given by lab bhattacharjee But unfortunately I couldn't find its proof.Can you provide a link Where the result you mentioned is proved? Because I have encountered this formula for the first time And It just seems like magic to me. – Freelancer Mar 21 '16 at 13:01
  • Also should I delete the question because it turns out that by using the result mentioned by crostul .the question Becomes Trivial.And the answer is clearly $a=4$...so This question becomes same as asking what is $2+2$...and so I don't think it will be of any help to anyone in the future. – Freelancer Mar 21 '16 at 13:04
  • Unfortunately I have no proof at hand. – Crostul Mar 21 '16 at 14:04

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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 : $$D=a,\quad E=4,\quad F=1.$$

mathlove
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