# Axes-based equations of conics

Every parabola can be represented in the general form $$\text{Axis}^2= \text{LR} \cdot (\text{Tangent})$$ where the tangent is at the vertex and LR is the latus rectum.

1. How is this derived?
2. Are there similar expressions for ellipses and hyperbolas using the axes? If so, how are they derived?

Update: It seems that replacing $$x$$ and $$y$$ in their standard equations with the distance from each axis gives the general equation, but I can't derive these. Can somebody help out?

• "How is this derived?" To answer this, one should know what starting point you have in mind. Jun 23 at 16:36
• You might want to be more specific of how to read that equation. My initial interpretation was along the lines of math.stackexchange.com/q/3135947/35416 but reading the answer there told me you're probably taking about distances not polynomials for axis and tangent.
– MvG
Jun 23 at 18:17
• @MvG: no, it's the equations that I meant, not the distances. Jun 24 at 1:20
• @Intelligentipauca: I would guess the basic definition of the conic with a focus and directrix. Jun 24 at 1:22
• With focus $(f_1,f_2)$ and directrix $ax+by+c=0$ the definition of the parabola translates to $$(x-f_1)^2+(y-f_2)^2-(ax+by+c)^2/(a^2+b^2))=0.$$ You want the form $$\frac{(ay-bx-af_2+bf_1)^2}{a^2+b^2}-(2\frac{bf_2+af_1+c}{\sqrt{a^2+b^2}})\frac{(ax+by-\frac{bf_2+af_1-c}{2})}{\sqrt{a^2+b^2}}=0$$ which can be proved is the same by expanding both forms and comparing. Aug 3 at 7:58

Suppose point $$P$$ on a parabola with focus $$F$$ and latus rectum $$4r$$ has distance $$p$$ from the axis and distance $$q$$ from the vertex-tangent.

Then $$\overline{PF}$$ is the hypotenuse of a right triangle with legs $$p$$ and $$|q-r|$$; also, $$P$$ is at distance $$q+r$$ from the directrix. These distances are equal for a parabola, so we can equate their squares:

$$p^2+|q-r|^2 = (q+r)^2 \qquad\to\qquad p^2=4r\cdot q \tag{1}$$

Now, we simply observe that the (signed) distance from a point $$(x,y)$$ to a line with equation $$ax+by+c=0$$ is given by the normalized form of the equations left-hand side; ie, $$\text{distance to line} = \frac{ax+by+c}{\sqrt{a^2+b^2}} \tag{2}$$ so we can write $$(1)$$ as $$(\text{normalized eqn of axis})^2 = (\text{length of latus rectum})\cdot(\text{normalized eqn of vertex-tangent}) \tag{1'}$$ as desired. $$\square$$

Importantly, the equation of a line has a sign ambiguity (eg, $$1x+2y-3=0$$ represents the same line as $$-1x-2y+3=0$$). In the context of $$(1')$$, the ambiguity with the axis equation is rendered irrelevant because of squaring; however, the ambiguity with the vertex-tangent equation corresponds to the parabola opens. That is to say: a given parabola requires an unambiguous vertex-tangent equation; namely, $$Ax+By+C=0$$ where normal vector $$(A,B)$$ points in the direction that the parabola opens.

To generalize to other conics, recall that a point on a conic eccentricity $$e$$ satisfies $$(\text{dist to focus})=\text{eccentricity}\cdot(\text{dist to directrix})\tag{3}$$ It is "known" that, for a central conic with major/transverse radius $$a$$, the center-to-focus distance is $$c=ae$$, and the center-to-directrix distances is $$d:=a/e$$; therefore, we can calculate the vertex-to-focus and vertex-to-directrix distances as $$|VF|=\sigma(a-c)=\sigma a(1-e) \qquad\qquad |VD|=\sigma(d-a)=\sigma\frac{a}{e}(1-e) \tag4$$ where $$\sigma$$ is $$+1$$ for ellipses and $$-1$$ for hyperbolas. With these, the square of relation $$(3)$$ gives this analogue of $$(1)$$: $$p^2+|q-|VF||^2 = e^2(p+|VD|)^2 \quad\to\quad p^2 + q^2(1-e^2)=2\sigma a (1-e^2) \cdot q \tag5$$ Now, note that if the minor/conjugate radius of the conic is $$b$$, then $$b^2=\sigma(a^2-c^2)=\sigma a^2(1-e^2)$$, so that we can write $$(5)$$ as $$p^2 + q^2(1-e^2)=\frac{2b^2}{a}\cdot q \tag6$$ But $$2b^2/a$$ is precisely the length of the latus rectum, so this gives us the analogue of $$(1')$$:

$$(\text{axis eqn})^2+(\text{tangent eqn})^2(1-\text{eccentricity}^2)=(\text{latus rectum})(\text{tangent eqn}) \tag{7}$$

where the equations are normalized, as before. $$\square$$

• The second part of your answer, if I understand it correctly, implies that the normal vector $(a, b)$ of any line $ax+by+c=0$ points into the half-plane where the points have an algebraic distance to the line that is positive. Could you explain how that's derived? Splendid answer, btw. Aug 10 at 9:03
• @harry: (Thanks for the "splendid". :) ... Suppose $P=(p,q)$ lies on the line $ax+by+c=0$. (For simplicity we'll assume $a^2+b^2=1$; if this isn't so already, we can always divide-through by $\sqrt{a^2+b^2}$ to make it so.) Thus, $ap+bq+c=0$. Translating $P$ by signed distance $d$ relative to normal direction $(a,b)$ gets us to point $(p+ad,q+bd)$. We observe $$a(p+ad)+b(q+bd)+c=(ap+bq+c)+d(a^2+b^2)=d$$ Thus, $ax+by+c$ is itself the signed-distance-to-the-line function that naturally considers distances on the normal-directed side of the line to be positive.
– Blue
Aug 10 at 9:32
• Isn't it correct only when we consider d to be a magnitude of the distance travelled, and not the signed distance? It seems to work that way. If it is so, I got it, thanks. (For anyone else who didn't get it at first either, @Blue used the polar form to represent the point in the half-plane; the sine and cosine usually used is present as a and b, which are the respective directional cosine and sine of the normal vector.) Aug 10 at 11:08
• @harry: Perhaps my comment could have been clearer, but that's why I wrote that $d$ was a signed distance "relative to normal direction $(a,b)$". By that I meant $d$ is considered positive when $P$ is translated in the direction of the normal, and negative when $P$ is translated in the opposite direction (and, of course, zero when $P$ isn't translated at all). This conveniently matches the $ax+by+c$ calculation, making for a consistent definition of "signed distance" from the line.
– Blue
Aug 10 at 11:30
• That clears it all up, thanks. Aug 10 at 11:32