Differential equation of a conic with given constraints

Question

The differential equation of all conics whose axes coincide with the axes of coordinates is of order

(A) 2

(B) 3

(C) 4

(D) 1

I know that the general equation of a conic is:

If I apply Lagrange multiplier theorem I get:

How should I continue with this problem?

Answer 1

The equations that you got are those lines which are parallel to the axes. In that case, either of $a,b$ or $h$ must be zero. But $a,b$ can't be zero (the curve won't be quadratic then). So $h=0$. Also the axes coincide with the coordinate axes. So, applying this one more constraint gives $g=f=0$. That means, the equation of all such conics is $ax^2+by^2+c=0$ or $x^2+\dfrac ba y^2+ \dfrac ca=0$. Two arbitrary constants, so second order.

EDIT:

$$x^2+\dfrac ba y^2+ \dfrac ca=0$$

Differentiating w.r.t $x$ gives:

$$2x+2\dfrac ba y\cdot y'=0$$

or: $$\dfrac x {y\; y'}+\dfrac ba =0$$

Another differentiation gets us rid of all the arbitrary constants, and we end up with a $2$nd order differential equation.