Parabola in Coordinate Geometry
Parabola: Definition based on Focus and Directrix
The parabola is one of the three non-degenerate conic sections (along with the ellipse and hyperbola). Beyond its formation by intersecting a cone with a plane parallel to a generator, a parabola can be defined as a locus of points based on distances to a fixed point and a fixed line. This definition provides a clear geometric property that characterizes all points on a parabola.
Definition
A parabola is defined as the locus of all points $P$ in a plane that are equidistant from a fixed point $F$ and a fixed straight line $L$ in the same plane. The fixed point $F$ is called the focus, and the fixed line $L$ is called the directrix. A crucial part of the definition is that the focus $F$ must not lie on the directrix $L$.

Let $P(x, y)$ be any point on the parabola.
Let $F$ be the focus (a fixed point).
Let $L$ be the directrix (a fixed line).
Let $M$ be the foot of the perpendicular drawn from the point $P$ to the line $L$. Then $PM$ is the perpendicular distance from $P$ to the directrix $L$.
According to the definition, the defining condition for any point $P$ to be on the parabola is that the distance from $P$ to the focus is equal to the perpendicular distance from $P$ to the directrix:
$\mathbf{\text{Distance from P to F} = \text{Perpendicular distance from P to L}}$
$\mathbf{|PF| = |PM|}$
This definition is consistent with the general focus-directrix definition of conic sections, $\frac{|PF|}{|PM|} = e$, where the eccentricity $e$ is equal to 1 for a parabola ($e = 1$).
Key Terminology from Definition
Based on the focus-directrix definition, several key features of a parabola can be identified:
-
Focus (F): The fixed point used in the definition ($|PF| = |PM|$). The shape of the parabola "opens up" around the focus.
-
Directrix (L): The fixed straight line used in the definition. The parabola curves away from the directrix.
-
Axis of Symmetry (or Axis): The straight line that passes through the focus $F$ and is perpendicular to the directrix $L$. The parabola is perfectly symmetric with respect to this line. For every point P on the parabola, its reflection across the axis is also on the parabola.
-
Vertex (V): The point where the parabola intersects its axis of symmetry. This point is unique and lies exactly midway between the focus and the directrix. It is the point on the parabola that is closest to the directrix (and hence closest to the focus). By the definition $|VF| = |VM|$, and since V is on the axis which is perpendicular to the directrix, VM is the perpendicular distance from V to the directrix. V is also the turning point of the parabola's curve.

Understanding these terms is essential for working with the standard equations and properties of parabolas.
Parabola Definition (Summary)
Definition:
Locus of points P such that distance from P to Focus F equals perpendicular distance from P to Directrix L.
$\mathbf{|PF| = |PM|}$
Eccentricity:
$\mathbf{e = 1}$
Key Terms:
- Focus (F): Fixed point.
- Directrix (L): Fixed line (F not on L).
- Axis: Line through F, perpendicular to L.
- Vertex (V): Point on the parabola where it crosses the axis; midway between F and L.
Key Property:
Equidistance from focus and directrix.
Standard Equations of Parabola ($y^2 = 4ax$, etc.)
The equation of a parabola takes its simplest form when the parabola is positioned in a standard orientation relative to the coordinate axes. These standard forms are obtained when the vertex of the parabola is located at the origin $(0, 0)$ and its axis of symmetry coincides with either the x-axis or the y-axis.
There are four such standard orientations, resulting in four standard equations for a parabola with vertex at the origin. We will derive the equation for one case and then summarize the others.
Case 1: Axis along the x-axis, Vertex at Origin, Opening Right ($y^2 = 4ax$)
Let the vertex of the parabola be at the origin $V(0, 0)$.
Let the axis of symmetry be the x-axis (whose equation is $y=0$).
Assume the parabola opens to the right, which means the focus lies on the positive x-axis and the directrix is a vertical line to the left of the origin.
Let the distance from the vertex to the focus be $a$. Since the vertex is at the origin and the axis is the x-axis, the focus $F$ has coordinates $(a, 0)$. For the parabola to open right, $a$ must be a positive value ($a > 0$).
By the definition of a parabola, the vertex V is equidistant from the focus F and the directrix L. The distance $|VF| = a$. Since the axis is perpendicular to the directrix, the directrix must be a vertical line. As V(0,0) is the midpoint between F(a,0) and the directrix, the directrix must be the line $x = -a$, or $x + a = 0$.

Derivation using $|PF| = |PM|$:
Let $P(x, y)$ be any point on the parabola.
The distance from $P(x, y)$ to the focus $F(a, 0)$ is given by the distance formula:
$|PF| = \sqrt{(x - a)^2 + (y - 0)^2} = \sqrt{(x - a)^2 + y^2}$
... (i)
The perpendicular distance from the point $P(x, y)$ to the directrix $L: x + a = 0$ is given by the distance from a point to a line formula, $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Here, $(x_0, y_0) = (x, y)$, $A=1, B=0, C=a$.
$|PM| = \frac{|1(x) + 0(y) + a|}{\sqrt{1^2 + 0^2}} = \frac{|x + a|}{\sqrt{1}} = |x + a|$
... (ii)
Alternatively, the foot of the perpendicular from $P(x, y)$ to the vertical line $x=-a$ is the point $M(-a, y)$. The distance $|PM|$ is the distance between $(x, y)$ and $(-a, y)$: $|PM| = \sqrt{(x - (-a))^2 + (y - y)^2} = \sqrt{(x+a)^2 + 0} = \sqrt{(x+a)^2} = |x+a|$.
By the definition of the parabola, $|PF| = |PM|$.
$\sqrt{(x - a)^2 + y^2} = |x + a|$
($|PF| = |PM|$)
To eliminate the square root and absolute value, square both sides of the equation:
$(\sqrt{(x - a)^2 + y^2})^2 = (|x + a|)^2$
... (iii)
$(x - a)^2 + y^2 = (x + a)^2$
... (iv)
Expand the squared terms in equation (iv):
$x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2$
... (v)
Cancel $x^2$ and $a^2$ from both sides of equation (v):
$-2ax + y^2 = 2ax$
... (vi)
Move the $-2ax$ term to the right side of equation (vi):
$y^2 = 2ax + 2ax$
... (vii)
$\mathbf{y^2 = 4ax}$
... (viii)
Equation (viii) is the standard equation of a parabola with vertex at the origin, axis along the x-axis, focus at $(a, 0)$, and directrix $x = -a$. This form represents a parabola opening towards the positive x-direction when $a > 0$. If we consider the case where the focus is at $(-a, 0)$ with $a>0$, the directrix is $x=a$, and the derivation leads to $y^2 = -4ax$, representing a parabola opening to the left.
Summary of Standard Equations (Vertex at Origin)
Based on whether the axis of symmetry is the x-axis or the y-axis, and whether the parabola opens in the positive or negative direction along that axis, we have four standard forms for a parabola with its vertex at the origin $(0, 0)$. In these equations, the parameter '$a$' is conventionally taken as a positive real number ($a>0$), representing the distance from the vertex to the focus (and from the vertex to the directrix).
Equation | Opening Direction | Axis of Symmetry | Focus (F) | Directrix (L) |
---|---|---|---|---|
$y^2 = 4ax$ | Right (Positive x-direction) | x-axis ($y=0$) | $(a, 0)$ | $x = -a$ |
$y^2 = -4ax$ | Left (Negative x-direction) | x-axis ($y=0$) | $(-a, 0)$ | $x = a$ |
$x^2 = 4ay$ | Up (Positive y-direction) | y-axis ($x=0$) | $(0, a)$ | $y = -a$ |
$x^2 = -4ay$ | Down (Negative y-direction) | y-axis ($x=0$) | $(0, -a)$ | $y = a$ |
These standard forms are the basis for understanding the properties of parabolas and for analyzing parabolas whose vertex is not at the origin (by shifting the origin or using the standard form with translated coordinates).
Example 1. Find the equation of the parabola with vertex at the origin and focus at (3, 0).
Answer:
Given:
Vertex is at the origin $V(0, 0)$.
Focus is at $F(3, 0)$.
Since the vertex is at the origin and the focus is on the positive x-axis (at a distance of 3 units from the origin), the axis of symmetry is the x-axis, and the parabola opens to the right (in the positive x-direction).
The standard equation for a parabola with vertex at the origin and opening to the right is $\mathbf{y^2 = 4ax}$, where $a$ is the distance from the vertex to the focus.
The distance from the vertex $V(0, 0)$ to the focus $F(3, 0)$ is $a = \sqrt{(3 - 0)^2 + (0 - 0)^2} = \sqrt{3^2} = 3$.
$a = 3$
... (i)
Substitute the value of $a$ from equation (i) into the standard equation $y^2 = 4ax$:
$y^2 = 4(3)x$
... (ii)
Simplify equation (ii):
$\mathbf{y^2 = 12x}$
... (iii)
Equation (iii) is the required equation of the parabola.
From this equation, we can also find the directrix. The directrix is a vertical line $x = -a$. So, the directrix is $x = -3$.
Standard Parabola Equations (Summary)
Vertex at Origin (0,0), $|a|$ = distance V to F:
Equation | Opens | Axis | Focus | Directrix | Latus Rectum Length |
---|---|---|---|---|---|
$y^2 = 4ax$ ($a>0$) | Right | x-axis ($y=0$) | $(a, 0)$ | $x = -a$ | $4a$ |
$y^2 = -4ax$ ($a>0$) | Left | x-axis ($y=0$) | $(-a, 0)$ | $x = a$ | $4a$ |
$x^2 = 4ay$ ($a>0$) | Up | y-axis ($x=0$) | $(0, a)$ | $y = -a$ | $4a$ |
$x^2 = -4ay$ ($a>0$) | Down | y-axis ($x=0$) | $(0, -a)$ | $y = a$ | $4a$ |
General Vertex $(h, k)$:
Replace $x$ with $(x-h)$ and $y$ with $(y-k)$ in the standard equations. E.g., Vertex $(h, k)$, opening right: $(y-k)^2 = 4a(x-h)$.
Properties of Parabola (Focus, Directrix, Vertex, Axis, Latus Rectum)
Associated with each standard equation of a parabola are several key geometric features: the focus, directrix, vertex, axis of symmetry, and latus rectum. These elements help define the shape and position of the parabola and are important for understanding its properties.
For the standard equations with the vertex at the origin, these properties can be directly identified or calculated based on the parameter '$a$'. Recall that '$a$' is the distance from the vertex to the focus (and to the directrix), and it is conventionally taken as a positive value ($a>0$).
Latus Rectum
The latus rectum is a specific chord of the parabola that provides a measure of its width or "openness" at the focus. It is defined as the chord that passes through the focus, is perpendicular to the axis of symmetry, and has its two endpoints lying on the parabola.
Derivation of the Length of the Latus Rectum:
Let's consider the standard parabola $y^2 = 4ax$ (where $a > 0$), which opens to the right, has its vertex at $(0, 0)$, and its axis of symmetry along the x-axis. The focus of this parabola is $F(a, 0)$.
The latus rectum passes through the focus $(a, 0)$ and is perpendicular to the axis (the x-axis, $y=0$). A line perpendicular to the x-axis is a vertical line, with equation $x = \text{constant}$. Since it passes through $(a, 0)$, its equation is $x = a$.
To find the endpoints of the latus rectum, we need to find the points where the line $x = a$ intersects the parabola $y^2 = 4ax$. Substitute $x = a$ into the equation of the parabola:
$y^2 = 4a(a)$
... (i)
$y^2 = 4a^2$
... (ii)
Take the square root of both sides of equation (ii) to find the corresponding y-coordinates:
$y = \pm \sqrt{4a^2} = \pm 2a$
... (iii)
So, when $x = a$, the y-coordinates are $2a$ and $-2a$. The endpoints of the latus rectum are $L_1(a, 2a)$ and $L_2(a, -2a)$.

The length of the latus rectum is the distance between its endpoints $L_1(a, 2a)$ and $L_2(a, -2a)$. Since these points have the same x-coordinate, the distance is the absolute difference of their y-coordinates:
Length of Latus Rectum $= |2a - (-2a)| = |2a + 2a| = |4a|$
... (iv)
Since $a > 0$, $|4a| = 4a$.
Length of Latus Rectum $= 4a$
... (v)
The length of the latus rectum is $4a$. This is a characteristic property of parabolas and is always equal to the magnitude of the coefficient of the linear term in the standard equation ($y^2 = 4ax$ or $x^2 = 4ay$).
Summary of Properties for Standard Parabolas (Vertex at Origin)
The following table summarizes the key properties for the four standard forms of a parabola with its vertex at the origin. In this table, $a$ is always taken as a positive value ($a>0$), representing the distance from the vertex to the focus.
Property | $y^2 = 4ax$ | $y^2 = -4ax$ | $x^2 = 4ay$ | $x^2 = -4ay$ |
---|---|---|---|---|
(Applies for $a>0$) | (Opens Right) | (Opens Left) | (Opens Up) | (Opens Down) |
Vertex (V) | $(0, 0)$ | $(0, 0)$ | $(0, 0)$ | $(0, 0)$ |
Focus (F) | $(a, 0)$ | $(-a, 0)$ | $(0, a)$ | $(0, -a)$ |
Equation of Directrix (L) | $x = -a$ | $x = a$ | $y = -a$ | $y = a$ |
Equation of Axis | $y = 0$ (x-axis) | $y = 0$ (x-axis) | $x = 0$ (y-axis) | $x = 0$ (y-axis) |
Equation of Latus Rectum (Line containing it) | $x = a$ | $x = -a$ | $y = a$ | $y = -a$ |
Endpoints of Latus Rectum | $(a, 2a)$ and $(a, -2a)$ | $(-a, 2a)$ and $(-a, -2a)$ | $(2a, a)$ and $(-2a, a)$ | $(2a, -a)$ and $(-2a, -a)$ |
Length of Latus Rectum | $4a$ | $4a$ | $4a$ | $4a$ |
Eccentricity (e) | 1 | 1 | 1 | 1 |
These properties are essential for sketching the graph of a parabola and for solving problems involving its geometric features.
Example 1. For the parabola $x^2 = -16y$, find the coordinates of the focus, the equation of the directrix, the equation of the axis, and the length of the latus rectum.
Answer:
The given equation of the parabola is $x^2 = -16y$.
We compare this equation with the standard forms of a parabola with the vertex at the origin. The equation is of the form $x^2 = \text{constant} \times y$, which indicates that the axis of symmetry is the y-axis and the vertex is at the origin (since there are no $(x-h)^2$ or $(y-k)^2$ terms).
Specifically, the equation $x^2 = -16y$ matches the standard form $x^2 = -4ay$. This form represents a parabola with vertex at $(0, 0)$, axis along the y-axis, and opening downwards (due to the negative sign).
Comparing the coefficients of $y$ in $x^2 = -16y$ and $x^2 = -4ay$:
$-4a = -16$
... (i)
Solve equation (i) for the parameter $a$ (which is always taken as positive):
$4a = 16$
$a = \frac{16}{4} = 4$
... (ii)
Now that we have the value of $a$, we can use the properties of the standard form $x^2 = -4ay$ with $a=4$ to find the required features:
- Coordinates of the Focus (F): For $x^2 = -4ay$, the focus is $(0, -a)$. Substitute $a=4$:
Focus = $(0, -4)$.
- Equation of the Directrix (L): For $x^2 = -4ay$, the equation of the directrix is $y = a$. Substitute $a=4$:
Directrix: $y = 4$.
- Equation of the Axis: For $x^2 = -4ay$, the axis of symmetry is the y-axis, whose equation is $x = 0$.
Axis: $x = 0$.
- Length of the Latus Rectum: The length of the latus rectum for any standard parabola is $4a$. Substitute $a=4$:
Length of Latus Rectum = $4(4) = 16$.
Summary of the properties for the parabola $x^2 = -16y$:
Coordinates of the focus: $\mathbf{(0, -4)}$.
Equation of the directrix: $\mathbf{y = 4}$.
Equation of the axis: $\mathbf{x = 0}$.
Length of the latus rectum: $\mathbf{16}$.
The parabola opens downwards from the vertex (0,0), its focus is 4 units below the vertex on the y-axis, and its directrix is a horizontal line 4 units above the vertex.
Parabola Properties (Summary)
For Standard Forms (Vertex at Origin):
See the table provided earlier for properties like Focus, Directrix, Axis, Latus Rectum for $y^2=\pm 4ax$ and $x^2=\pm 4ay$ (with $a>0$).
Latus Rectum:
Chord through Focus, perpendicular to Axis, endpoints on parabola.
Length of Latus Rectum $= \mathbf{4a}$ (for standard equations $y^2=\pm 4ax$ or $x^2=\pm 4ay$, where $4a$ is the magnitude of the coefficient of the linear term).
Eccentricity:
Always $\mathbf{e=1}$ for a parabola.
Vertex $(h, k)$:
If the vertex is at $(h, k)$, the standard equations are translated: e.g., $(y-k)^2 = 4a(x-h)$, Focus $(h+a, k)$, Directrix $x = h-a$, Axis $y=k$, Latus Rectum endpoints $(h+a, k \pm 2a)$, Length $4a$. Apply similar translations for other orientations.
Parametric Equations of Parabola
While Cartesian equations (like $y^2 = 4ax$) define a relationship between $x$ and $y$ coordinates, parametric equations define both $x$ and $y$ in terms of a third independent variable, called a parameter. Parametric representations are useful for tracing the curve and analyzing motion along the curve, and they can sometimes simplify calculations involving points on the parabola.
Parametric Form for $y^2 = 4ax$
Consider the standard parabola with equation $y^2 = 4ax$. We want to find expressions for $x$ and $y$ in terms of a parameter, say $t$, such that for every value of $t$, the point $(x(t), y(t))$ lies on the parabola, and as $t$ varies, all points on the parabola are covered.
A common choice for the parameterization of $y^2 = 4ax$ involves setting one of the coordinates or a simple expression in terms of the parameter. A convenient approach is to look at the linear term $4ax$ and the squared term $y^2$. If we make a substitution involving $y$, like $y = ct$ for some constant $c$, then $y^2$ becomes $c^2 t^2$. We need $x$ to be proportional to $t^2$ to match the structure $y^2 \propto x$.
Let's try setting $y = 2at$. Substituting this into the parabola equation $y^2 = 4ax$:
$(2at)^2 = 4ax$
... (i)
Simplify equation (i):
$4a^2 t^2 = 4ax$
... (ii)
Assuming $a \neq 0$ (for a non-degenerate parabola) and $x \neq 0$ (which is true for points other than the vertex), we can divide both sides by $4a$:
$\frac{4a^2 t^2}{4a} = \frac{4ax}{4a}$
... (iii)
$at^2 = x$
... (iv)
So, if $y = 2at$, then $x = at^2$. We can verify that the point $(at^2, 2at)$ satisfies $y^2 = 4ax$ for any real value of $t$: $(2at)^2 = 4a^2 t^2$ $4a(at^2) = 4a^2 t^2$ The equation is satisfied.
As the parameter $t$ varies over all real numbers, the point $(at^2, 2at)$ sweeps out the entire parabola $y^2 = 4ax$. For instance, if $t=0$, the point is $(0,0)$ (the vertex). As $t$ increases, $x$ increases ($at^2$), and $|y|$ increases ($|2at|$), tracing the upper and lower branches. As $t \to \infty$, $x \to \infty$ and $|y| \to \infty$. As $t \to -\infty$, $x \to \infty$ and $y \to -\infty$.
The standard parametric equations for the parabola $y^2 = 4ax$ are:
$\mathbf{x = at^2}$
$\mathbf{y = 2at}$
The point $(at^2, 2at)$ is often referred to simply as the point 't' on the parabola $y^2 = 4ax$.
Parametric Forms for Other Standard Parabolas
Similar parametric equations can be found for the other standard orientations of the parabola with vertex at the origin. The general idea is to ensure that the square of one coordinate (or its proportional expression) is proportional to the other coordinate (or its proportional expression), mimicking the form of the Cartesian equation.
Cartesian Equation | Parametric Equations |
---|---|
$y^2 = 4ax$ ($a>0$) | $x = at^2, \quad y = 2at$ |
$y^2 = -4ax$ ($a>0$) | $x = -at^2, \quad y = 2at$ (or $y=-2at$) |
$x^2 = 4ay$ ($a>0$) | $x = 2at, \quad y = at^2$ (or $x=-2at$) |
$x^2 = -4ay$ ($a>0$) | $x = 2at, \quad y = -at^2$ (or $x=-2at$) |
Note that the parameter $a$ in the parametric equations is the same positive value as in the standard Cartesian equations. The parameter $t$ can take any real value.
Parametric Form for General Parabola (Vertex $(h, k)$):
If the vertex of the parabola is at $(h, k)$, the parametric equations are obtained by translating the standard parametric equations. For example, for the parabola $(y-k)^2 = 4a(x-h)$, which opens right, the parametric equations are:
$x - h = at^2 \implies x = h + at^2$
$y - k = 2at \implies y = k + 2at$
Similar translations apply to the other orientations.
Example 1. Find the Cartesian equation of the curve whose parametric equations are $x = 3t^2$, $y = 6t$.
Answer:
Given the parametric equations:
$x = 3t^2$
... (1)
$y = 6t$
... (2)
To find the Cartesian equation, we need to eliminate the parameter $t$ and find a direct relationship between $x$ and $y$. We can solve one equation for $t$ and substitute it into the other.
From the simpler equation (2), solve for $t$:
$t = \frac{y}{6}$
... (iii)
Now, substitute this expression for $t$ from equation (iii) into equation (1):
$x = 3 \left( \frac{y}{6} \right)^2$
... (iv)
Simplify equation (iv):
$x = 3 \left( \frac{y^2}{6^2} \right) = 3 \left( \frac{y^2}{36} \right)$
$x = \frac{3y^2}{36}$
... (v)
Simplify the fraction in equation (v):
$x = \frac{y^2}{12}$
... (vi)
Rearrange equation (vi) to match a standard form:
$\mathbf{y^2 = 12x}$
... (vii)
Equation (vii) is the Cartesian equation of the curve. This is the standard equation of a parabola opening to the right with vertex at the origin. Comparing it with $y^2 = 4ax$, we have $4a = 12$, so $a = 3$. The parametric equations $x=3t^2, y=6t$ are indeed of the form $x=at^2, y=2at$ with $a=3$.
Parabola Parametric Equations (Summary)
For Standard Parabola $y^2 = 4ax$ ($a>0$):
$\mathbf{x = at^2, \quad y = 2at}$
For Other Standard Parabolas (Vertex at Origin, $a>0$):
- $y^2 = -4ax$: $x = -at^2, y = 2at$
- $x^2 = 4ay$: $x = 2at, y = at^2$
- $x^2 = -4ay$: $x = 2at, y = -at^2$
For Parabola with Vertex $(h, k)$ (e.g., $(y-k)^2 = 4a(x-h)$):
$\mathbf{x = h + at^2, \quad y = k + 2at}$
Key Idea:
Represent points on the curve using a single parameter $t$. Useful for analysis and sometimes simplifies problem-solving.
Converting from Parametric to Cartesian:
Eliminate the parameter $t$ by solving one equation for $t$ and substituting into the other.