Mathematics · Ch 10 — Conic Sections

Standard Equations of Parabola

10.4.1

Standard Equations of Parabola

The Standard Equations of a Parabola

The simplest and most useful form of a parabola's equation occurs when its vertex is placed at the origin and its axis of symmetry lies along either the x-axis or the y-axis. There are four possible orientations for such a parabola, depending on which direction it opens: right, left, upward, or downward. We will derive the equation for each orientation, starting with the parabola that opens to the right.

Deriving the Equation: Parabola Opening to the Right

Consider a parabola with its focus at $(a, 0)$ where $a > 0$ , and its directrix as the vertical line $x = -a$ . The vertex is at the origin $(0,0)$ .

Let $F$ be the focus and $l$ the directrix. Draw a perpendicular $FM$ from $F$ to $l$ . The midpoint $O$ of $FM$ lies on the parabola by definition (it is equidistant from $F$ and $l$ ) and is called the vertex. We set $O$ as the origin, $OX$ as the x-axis (along $FM$ ), and $OY$ as the y-axis (perpendicular to $OX$ ). The distance from the directrix to the focus is $2a$ , so the focus is at $(a, 0)$ and the directrix is the line $x + a = 0$ .

Now, let $P(x, y)$ be any point on the parabola. By the definition of a parabola, the distance from $P$ to the focus equals the distance from $P$ to the directrix. The perpendicular distance from $P$ to the directrix is the distance to the point $B(-a, y)$ on the directrix directly above or below $P$ .

Therefore, the condition is:

PF = PB

Using the distance formula:

PF = \sqrt{(x - a)^2 + y^2}

PB = \sqrt{(x + a)^2}

Setting them equal:

\sqrt{(x - a)^2 + y^2} = \sqrt{(x + a)^2}

Squaring both sides:

(x - a)^2 + y^2 = (x + a)^2

Expanding:

x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2

Cancelling $x^2$ and $a^2$ from both sides:

-2ax + y^2 = 2ax

Rearranging gives the equation of the parabola:

y^2 = 4ax \quad (a > 0)

›Proof

Proving the Converse

We must also show that any point $P(x, y)$ satisfying $y^2 = 4ax$ lies on the parabola. Calculate $PF$ :

$PF = \sqrt{(x - a)^2 + y^2} = \sqrt{(x - a)^2 + 4ax}$

$= \sqrt{x^2 - 2ax + a^2 + 4ax} = \sqrt{x^2 + 2ax + a^2} = \sqrt{(x + a)^2}$

Since $x \ge 0$ for this parabola (as we will see), $x + a > 0$ , so $\sqrt{(x + a)^2} = x + a = PB$ . Hence $PF = PB$ , and $P$ lies on the parabola.

Thus, the equation $y^2 = 4ax$ is the standard equation of a parabola with vertex at the origin, focus at $(a, 0)$ , and directrix $x = -a$ .

Note

Discussion of the Equation $y^2 = 4ax$

Since $a > 0$ , the term $4ax$ is non-negative only when $x \ge 0$ . This means $x$ can be any positive number or zero, but never negative. The curve therefore lies entirely in the first and fourth quadrants, extending infinitely to the right. The axis of symmetry is the positive x-axis.

The Four Standard Equations

By similar derivations, we obtain the equations for the other three orientations. The results are:

Orientation	Focus	Directrix	Standard Equation
Opens to the right	$(a, 0)$	$x = -a$	$y^2 = 4ax$
Opens to the left	$(-a, 0)$	$x = a$	$y^2 = -4ax$
Opens upward	$(0, a)$	$y = -a$	$x^2 = 4ay$
Opens downward	$(0, -a)$	$y = a$	$x^2 = -4ay$

In all cases, $a > 0$ . These four equations are known as the standard equations of parabolas.

Important

Key Feature of Standard Equations

In every standard equation, the focus lies on one coordinate axis, the vertex is at the origin, and the directrix is parallel to the other coordinate axis.

Observations from the Standard Equations

From the standard equations, we can make three important observations about the parabola's geometry.

1. Symmetry

A parabola is symmetric with respect to its own axis.

If the equation contains a $y^2$ term (like $y^2 = 4ax$ or $y^2 = -4ax$ ), the axis of symmetry is the x-axis.
If the equation contains an $x^2$ term (like $x^2 = 4ay$ or $x^2 = -4ay$ ), the axis of symmetry is the y-axis.

2. Direction of Opening (Axis along x-axis)

When the axis of symmetry is the x-axis: …

Figure 10.15Four standard parabolas (a)–(d)

Fig. 10.15 — Four standard parabolas (a)–(d)

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

The figure shows four separate X-Y coordinate grids, labelled (a) through (d). Each grid has the origin at the centre, with the x-axis horizontal and the y-axis vertical. In every panel, the vertex of the parabola is at the origin, and the axis of symmetry lies along one of the coordinate axes. A blue vertical or horizontal line marks the directrix, and a single point (the focus) is placed on the axis of symmetry at a distance $a$ from the vertex.

Panel (a) shows the parabola $y^2 = 4ax$ . The focus is at $(a,0)$ on the positive x-axis, and the directrix is the vertical line $x = -a$ (shown in blue). The curve opens to the right, lying entirely in the first and fourth quadrants (x is never negative). The axis of symmetry is the x-axis itself.

Panel (b) shows $y^2 = -4ax$ . The focus is at $(-a,0)$ on the negative x-axis, and the directrix is $x = a$ . The parabola opens to the left, with x never positive. The axis is still the x-axis.

Panel (c) shows $x^2 = 4ay$ . The focus is at $(0,a)$ on the positive y-axis, and the directrix is the horizontal line $y = -a$ . The parabola opens upward, with y never negative. The axis of symmetry is the y-axis.

Panel (d) shows $x^2 = -4ay$ . The focus is at $(0,-a)$ on the negative y-axis, and the directrix is $y = a$ . The parabola opens downward, with y never positive.

The physical idea the figure teaches is that a parabola is the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix). By placing the vertex at the origin and aligning the axis with a coordinate axis, the equation becomes simple. The sign of the coefficient tells you which direction the parabola opens: positive coefficient of $x$ in $y^2 = 4ax$ means right; negative means left. For $x^2 = 4ay$ , positive coefficient of $y$ means up; negative means down.

y^2 = 4ax \quad (a>0)

The other three standard forms follow by replacing $x$ with $-x$ or swapping $x$ and $y$ :

y^2 = -4ax \quad \text{(opens left, focus }(-a,0)\text{, directrix }x=a\text{)}

x^2 = 4ay \quad \text{(opens up, focus }(0,a)\text{, directrix }y=-a\text{)}

x^2 = -4ay \quad \text{(opens down, focus }(0,-a)\text{, directrix }y=a\text{)}

In every case, $a$ is the distance from the vertex to the focus (and also from the vertex to the directrix). The number $4a$ is called the latus rectum — the length of the chord through the focus perpendicular to the axis. …

Figure 10.16Derivation of y² = 4ax

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

The figure is a coordinate-plane diagram built around the definition of a parabola. The x‑axis and y‑axis are drawn, with the origin labelled O — this is the vertex of the parabola. The curve opens to the right, symmetric about the x‑axis. A point F is marked on the positive x‑axis at coordinates $(a,0)$ ; this is the focus. A vertical line labelled $l$ is drawn to the left of the origin, with equation $x + a = 0$ (equivalently $x = -a$ ); this is the directrix. The distance from the directrix to the focus is $2a$ , so the vertex O lies exactly midway between them.

A general point $P(x,y)$ on the parabola is shown. From P, a perpendicular segment is dropped to the directrix, meeting it at point $B(-a,y)$ . The foot of the perpendicular from the focus to the directrix is labelled $M(-a,0)$ . Two line segments are drawn: $PF$ (from P to the focus) and $PB$ (from P to the directrix). The entire diagram is the geometric statement of the parabola’s definition: for every point P on the curve, the distance to the focus equals the distance to the directrix.

The physical idea is that a parabola is the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix). The figure makes this equality visible: $PF$ and $PB$ are the two distances being compared. By placing the vertex at the origin and the axis along the x‑axis, the algebra becomes clean.

The textbook uses this figure to derive the standard equation. Starting from the definition $PF = PB$ , and using the coordinates $F(a,0)$ , $B(-a,y)$ , and $P(x,y)$ , the distance formula gives:

PF = \sqrt{(x-a)^2 + y^2}, \qquad PB = \sqrt{(x+a)^2}

Setting them equal and squaring:

(x-a)^2 + y^2 = (x+a)^2

Expanding both sides:

x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2

Cancelling $x^2$ and $a^2$ leaves:

y^2 = 4ax

y^2 = 4ax \quad (a > 0)

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