We can write general equation of conic as:

$$ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{a^2(1-e^2)} = 1$$

Where $a$ is some parameter and $e$ is eccentricity of conic

For e=0, it is a circle:

$$(x-h)^2 + (y-k)^2 = a^2$$


$0 < e<1$, is it an ellipse

$ e>1$, it is a hyperbola

Now, I want to derive the equation of parabola from this , where $e=1$, however that leads me to blowing up the expression. So, I isolated the expression for $e$:

$$ \frac{(y-k)^2}{ a^2 - (x-h)^2} = 1-e^2$$

$$ e^2 = 1 - \frac{(y-k)^2}{a^2 - (x-h)^2}$$

If we send $e \to 1$, this equation becomes:

$$(y-k)^2 =0$$

Which is the equation of a straight line... not a parabola. Why is that the equation didn't reduce to parabola?

  • $\begingroup$ Because you are not taking appropriate precautions when taking limits --- your focus and directrix into one another. $\endgroup$ – user10354138 Feb 24 at 9:14
  • $\begingroup$ Hmm how would I take such precautions @user10354138 $\endgroup$ – Buraian Feb 24 at 9:17
  • $\begingroup$ How do I know the coordinates of that point and line without having the equation of conic simplfiied out to give a curve? @user10354138 $\endgroup$ – Buraian Feb 24 at 9:26
  • $\begingroup$ What is general is the relation $PS=e PM$ where S(a,0) and Directrix is $x=-a$. You will get the parabola. In the ellipse $S(ae,0)$ and $x=a/e>0$ when you take $e=1$ the focus comes on directrix. So you cannot get parabola like this. $\endgroup$ – Z Ahmed Feb 24 at 10:22
  • $\begingroup$ Thank you, that helped @ZAhmed $\endgroup$ – Buraian Feb 24 at 11:13

What's wrong with your method: The ellipse/hyperbola has focus $(h+ c,k)$ and directrix $x=h+a^2/c$ where $c=\pm ae$. In the limit $e\to 1$, the foci crashes into the directrix so you end up with the perpendicular to the directrix through the foci.

To do it properly: You need to scale $a$ as a function of $e$ to to keep them apart.

Method 1: Fix the distance from vertex to directrix.

Fix the conic to have vertex at $(0,0)$ and the focus should be on the positive $x$-axis. So set the directrix at $x=-a$ and the focus at $(ae,0)$. The equation of the conic is $$ (x-ae)^2+y^2=e^2(x+a)^2 $$ Now let $e\to 1$ and you get $$ y^2=4ax $$ the equation of the parabola.

Method 2: Fix both the directrix and the focus. Let the focus be at $(a,0)$ and directrix $x=-a$. Then the equation of our conic is $$ (x-a)^2+y^2=e^2(x+a)^2 $$ Let $e\to 1$ gives $$ y^2=4ax $$ which isn't surprising.

  • $\begingroup$ Makes sense, but I wanted to make it using the original conci equation I wrote because from that one equation, we can understand all conic properties like focus for hyperbola,ellipse circle etc are given as :ae and directix as -a/e but, I guess the crash can't be avoided :( $\endgroup$ – Buraian Feb 24 at 11:23

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