On the Equations of Parabolas and Conic Sections

Conic sections

The parabola is a curve belonging to the family of conic sections, which are curves obtained by taking a cross-section of a cone.

Simulation · Visualising Conic Sections via Plane Tilt

You get different curves by tilting the plane: an ellipse (a scaled circle in two axes), a parabola (a single-branch curve), and a hyperbola (two branches).These are called conic sections

There is one property that binds all these curves together:

Given a point $F(x_{0},y_{0})\in \mathbb{R}²$ and a line (D) where F $\notin (D)$

If we consider the set of points $P(x,y) \in \mathbb{R}^2$ where the ratio of the distance to $F$ to the perpendicular distance to line $D$ is a constant, $e$.
This requirement generates all conic section curves.

Simulation · Locus of Points with a Constant PF:PD Ratio

This constant $e$ is called the eccentricity, and varying its value yields the different conic sections:

If $0 < e < 1$, we get an ellipse. Here, the distance from $P$ to $D$ is greater than the distance to $F$, but their ratio remains constant.

If $e = 1$, we get a parabola. Here, the distance from $P$ to $F$ equals the distance to $D$ ($PF = PD$).

If $e>1$ $\rightarrow$ we get a hyperbola.

Terminology

Point F is called the focus point and the line (D) is called the directrix.

The cartesian equation of the Parabola:

How do we derive the Cartesian equation of a parabola?
Let $P(x,y)$ be a general point on the parabola.
PF = PD ( with PD denoting the $\perp$ distance from P to line (D) )

We could keep everything completely general by using $F(x_0, y_0)$ and a general line $ax + by + c = 0$ for the directrix, but this unnecessarily complicates the distance calculations. Instead, we can assume a specific form for $F$ without losing any generality.
We will assume that point F is on the x axis, i.e, F could be written as $F(a,0)$

How can we assume $F$ lies on the $x$-axis without losing generality—in other words, why is the resulting equation invariant under such a transformation?1

We can rotate the coordinate axes to align $F$ with the $x$-axis.

This doesn't change the equation because $PF = PD$ involves distances, which are invariant under rotations.

Hence, making this assumption about $F$ does not result in any loss of generality.2

Simulation · show an x-y axies , and Show a point F and a line (D) with F not in D. then show how the x-y axes are rotated so that point F nowlies on the x axis and show then it's corrdinates becomes F(a,0)

Thus, we can set the focus as $F(a, 0)$. But what about the directrix?
It turns out it cannot be just any line—it must be perpendicular to the parabola's axis of symmetry.

How can we justify this?Let $F$ be the focus and $D$ be the directrix. The line passing through $F$ and perpendicular to $D$ forms a line of symmetry.

To see why, take a point $P$ on the parabola and reflect it across this axis to $P'$. Since reflection preserves distances along the axis, $PF = P'F$.Additionally, the perpendicular distance from $P'$ to the directrix $D$ equals that of $P$, so $P'D' = PD$ (where $D'$ is the perpendicular projection of $P'$ onto $D$).
Since $P$ lies on the parabola ($PF = PD$), it follows that $P'F = P'D'$. Thus, $P'$ also lies on the parabola, proving that this perpendicular line is a line of symmetry.

Simulation · Simulation

SO far, we have the focal point F on the x axis and we know that the directirx must be perpendicular to the line of symmetry i.e $\perp$ to the x axis.

There is one more assumption we can make to simplify the derivation without losing generality:

we can set the vertex $V$ at the origin $(0,0)$.
We achieve this by translating the axes so that the origin aligns with the vertex.

Why does this preserve the parabola's equation? Because the definition relies on distances. Translation simply shifts coordinates by a constant, which cancels out when computing distances—leaving the equation unchanged.

Simulation · Simulation

After all these justified assumptions we can now easily derive the cartesian equation of the parabola:

Simulation · The cartesian equation of a parabola

Finding the Equation of a Parabola When the Focus Is Not in Standard Position

Now, we know that If we are given a focus point in the form F(a,0) , then the equation of the parabola is given by $y^{2} = 4ax$

Student

But what if the focus isn't in that standard form—say, $F(3, 5)$ with a directrix like $y = 1$? How to get the equation of the parabola in such a case ?

Simulation · Transforming the standard parabola

We start by finding the vertex $V$, which always lies halfway between the directrix and the focus. Here, $V\left(3, 1 + \frac{5-1}{2}\right) = (3,3)$. Knowing $V$ tells us which direction the parabola opens (up, down, left, right), allowing us to sketch it—in this case, opening upward.

Next, consider a point $P(x,y)$ on the standard parabola (where $F(a,0)$ and directrix $x = -a$) and a point $P'(X',Y')$ on our new parabola. How do we map $P$ to $P'$? As seen in the simulation, we first rotate the standard parabola by 90° $$R\begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix}$$ This gives the standard parabola the correct orientation, but with a different vertex. We then translate the standard vertex to match our target parabola

$$\begin{pmatrix} X'\\ Y' \end{pmatrix}= R\begin{pmatrix} x\\ y \end{pmatrix} + \begin{pmatrix} 3\\ 3 \end{pmatrix} = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix} + \begin{pmatrix} 3\\ 3 \end{pmatrix}$$

$$\begin{cases} X'=y+3\\ Y'=-x+3 \end{cases}$$

$$\begin{cases} y=X'-3\\ x=3-Y' \end{cases}$$ Substituting back into the standard equation $y^2 = 4ax$, $a$ represents the distance from the vertex to the directrix, which is 2 in this case, we get:$$(X' - 3)^2 = 4(2)(3 - Y')$$. $$(X' - 3)^2 = -8 Y' +24$$

So, If the focus is not in standard form $F(a,0)$, what are the practical steps to find the Cartesian equation of the parabola?1

Locate $F$ and sketch the directrix $D$.

1) Find the vertex $V$ by calculating the midpoint between $F$ and $D$. This reveals the direction the parabola opens and gives the focal distance $a$

Relate a point $P(x,y)$ on the standard parabola to a point $P'(X',Y')$ on the target parabola via rotation and/or translation. Express this transformation in matrix form, then substitute into the standard equation to obtain the target equation.2