# Deriving equation of parabola from general equation of conic

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:

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

similarly,

$$0 < e<1$$, 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?

• Because you are not taking appropriate precautions when taking limits --- your focus and directrix into one another. – user10354138 Feb 24 at 9:14
• Hmm how would I take such precautions @user10354138 – Buraian Feb 24 at 9:17
• How do I know the coordinates of that point and line without having the equation of conic simplfiied out to give a curve? @user10354138 – Buraian Feb 24 at 9:26
• 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. – Z Ahmed Feb 24 at 10:22
• Thank you, that helped @ZAhmed – Buraian Feb 24 at 11:13

## 1 Answer

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.

• 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 :( – Buraian Feb 24 at 11:23