What is the condition for a second degree equation to represent a parabola?

Question

I have an idea about the answer to this , but not the clarity . Now the answer is $h^2=ab$ , if the equation is $ax^2+2hxy+by^2+2gx+2fy+c=0$ and that equation shouldn't be a pair of Straight lines. But how does the proof go. My attempt at it was take a line $lx+my+n=0$ as the directrix and $(h,k)$ as the focus and apply $e=1$ , But when I compare the equations I got , I am getting more than the conditions stated above . As there are three unknowns and three equations .

Answer 1

For parabola $h^2=ab$ and $\Delta= abc+2fgh-af^2-bg^2-ch^2\ne 0$

Answer 2

The special case $h=b=0$ and $a\ne0$ leads to the standard equation of a parabola having its axis parallel to $y$ axis. In the general case the axis of the parabola will have a slope $m$ and all lines parallel to the axis intersect the parabola at a single point (this is a peculiar property of the parabola).

Substituting $y=mx+k$ (equation of a line parallel to the axis) into the general equation of a conic $$ ax^2+2hxy+by^2+2gx+2fy+c=0 $$ we obtain $$ (bm^2+2hm+a)x^2+2(bkm+fm+g+hk)x+bk^2+2fk+c=0. $$ This equation can have a single solution for all values of $k$ only if the coefficient of $x^2$ vanishes, that is if: $$ bm^2+2hm+a=0. $$ For a parabola only a single value of $m$ is allowed, the slope of the axis, while in a hyperbola we have a single intersection for all lines parallel to either asymptote, hence two different values of $m$, and in an ellipse single intersections with all lines with a given slope are not possible. If the conic is a parabola, the discriminant of the above equation must then vanish, that is: $$ h^2-ab=0. $$ We get as a bonus the slope of the axis: $\displaystyle m=-{h\over b}$.

Answer 3

$y^2=4ax$ refers to the standard simple parabola where the axis $y=0$ is perpendicular to tan gent at vertex $x=0$. The length of latus rectum is $4a$.

So in a quadratic of $x,y$ if we can make perfect square and write it as $L_1^2=4AL_2$, where $L_1$ and $L_2$ are non parallel, it will be a parabola.

Hence, if we demand $$h^2=ab~~~~(1),$$ we can write $$(x \sqrt{a}+y \sqrt{b} y)^2=-2gx-2fy-c \implies L_1=x\sqrt{a}+y \sqrt{b} y, L_2=-2gx-2fy-c.$$ The condition that $m_1 \ne m_2$ gives $$f \sqrt{a}- g \sqrt{b} \ne 0~~~(2)$$ (1) and (2) when put in the expression of $\Delta$ give us $$\Delta=abc+2fgh-af^2-bg^2-ch^2=-[(f\sqrt{a}-g \sqrt{b}]^2 \ne 0~~~(3).$$

Normally, (1) and (3) are known to be necessary conditions for a parabola. (1) and (2) though equivalent are often overlooked.