Mathematics Parabola
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### Topics covered

star Definition
star Standard equations of parabola
star Latus rectum

### Definition

color{blue}(★ ul"Parabola")

\color{green} ✍️ A color{blue}(ul"parabola") is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane.

\color{green} ✍️ The fixed line is called the color{blue}(ul"directrix") of the parabola and the fixed point F is called the color{blue}("focus") (Fig \ \11.13).

\color{green} ✍️ color{blue}(ul"Para") means "for" and color{blue}(ul"bola") means "throwing", i.e., the shape described when you throw a ball in the air .

\color{green} ✍️ A line through the focus and perpendicular to the directrix is called the color{blue}(ul"axis of the parabola").

\color{green} ✍️ The point of intersection of parabola with the axis is called the color{blue}(ul"vertex of the parabola.") ( Fig \ \11.14).

color{blue} "☛ Note :" If the fixed point lies on the fixed line, then the set of points in the plane, which are equidistant from the fixed point and the fixed line is the straight line through the fixed point and perpendicular to the fixed line. We call this straight line as degenerate case of the parabola.

### Standard equations of parabola

\color{green} ✍️ The equation of a parabola is simplest if the vertex is at the origin and the axis of symmetry is along the x-axis or y-axis. The four possible such orientations of parabola are shown below inFig \ \ 11.15 (a) to (d).

\color{green} "We will derive the equation for the parabola shown above in Fig 11.15 (a)" with focus at (a, 0) a > 0; and directricx x = – a as below:

Let F be the focus and l the directrix. Let FM be perpendicular to the directrix and bisect FM at the point O. Produce MO to X.

By the definition of parabola, the mid-point O is on the parabola and is called the vertex of the parabola.

Take O as origin, OX the x-axis and OY perpendicular to it as the y-axis.

Let the distance from the directrix to the focus be 2a. Then, the coordinates of the focus are (a, 0), and the equation of the directrix is x + a = 0 as in Fig11.16. Let P(x, y) be any point on the parabola such that PF = PB,

where PB is perpendicular to l. The coordinates of B are (– a, y). By the distance formula, we have

PF = sqrt((x-a)^2 +y^2) and PB = sqrt((x+a)^2

Since PF = PB, we have

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

i.e. (x – a)^2 + y^2 = (x + a)^2

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

or y^2 = 4ax ( a > 0).

Hence, any point on the parabola satisfies

y^2 = 4ax.

Conversely, let P(x, y) satisfy the equation (2)

PF = sqrt((x-a)^2 +y^2) = sqrt((x-a)^2 + 4ax)

= sqrt[(x+a)^2] = PB

and so P(x,y) lies on the parabola.

Thus, from (2) and (3) we have proved that the equation to the parabola with vertex at the origin, focus at (a,0) and directrix x = – a is y^2 = 4ax.

Discussion In equation (2), since a > 0, x can assume any positive value or zero but no negative value and the curve extends indefinitely far into the first and the fourth quadrants. The axis of the parabola is the positive x-axis. Similarly, we can derive the equations of the parabolas in:

Fig 11.15 (b) as y^2 = – 4ax,
Fig 11.15 (c) as x^2 = 4ay,
Fig 11.15 (d) as x^2 = – 4ay,

These four equations are known as color{green}"standard equations of parabolas."

color{blue} "☛Note :" The standard equations of parabolas have focus on one of the coordinate axis; vertex at the origin and thereby the directrix is parallel to the other coordinate axis. However, the study of the equations of parabolas with focus at any point and any line as directrix is beyond the scope here.

### Latus rectum

color{blue} (★ ul"Latus rectum")

Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola (Fig11.17).

color{blue} (★ ul"To find the Length of the latus rectum of the parabola" \ \ y^2 = 4ax) (Fig 11.18).

By the definition of the parabola, AF = AC.

But AC = FM = 2a

Hence AF = 2a.

And since the parabola is symmetric with respect to x-axis AF = FB and so

AB = color{blue} (ul"Length of the latus rectum" = 4a.)

### Terms related to Parabola

Various terms related to parabola are explained in the image opposite.

### Different Types of Parabolas

color{blue}{ulmathtt"From the standard equations of the parabolas, we have the following observations:"}

color{blue}(☛) Parabola is symmetric with respect to the axis of the parabola.

If the equation has a y^2 term, then the axis of symmetry is along the x-axis and if the equation has an x^2 term, then the axis of symmetry is along the y-axis.

color{blue}(☛) When the axis of symmetry is along the x-axis the parabola opens to the
\ \ \ \ \ \ \ \ \ \(a) right if the coefficient of x is positive,
\ \ \ \ \ \ \ \ \ \(b) left if the coefficient of x is negative.

color{blue}(☛) When the axis of symmetry is along the y-axis the parabola opens
\ \ \ \ \ \ \ \ \ \(c) upwards if the coefficient of y is positive.
\ \ \ \ \ \ \ \ \ \(d) downwards if the coefficient of y is negative.
Q 3185691567

Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola y^2 = 8x.

Solution:

The given equation involves y^2, so the axis of symmetry is along the x-axis. The coefficient of x is positive so the parabola opens to the right. Comparing with the given equation y^2 = 4ax, we find that a = 2. Thus, the focus of the parabola is (2, 0) and the equation of the directrix of the parabola is x = – 2 (Fig 11.19). Length of the latus rectum is 4a = 4 × 2 = 8.
Q 3115691569

Find the equation of the parabola with focus (2,0) and directrix x = – 2.

Solution:

Since the focus (2,0) lies on the x-axis, the x-axis itself is the axis of the parabola. Hence the equation of the parabola is of the form either y^2 = 4ax or y^2 = – 4ax. Since the directrix is x = – 2 and the focus is (2,0), the parabola is to be of the form y^2 = 4ax with a = 2. Hence the required equation is y^2 = 4(2)x = 8x
Q 3125791661

Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2).

Solution:

Since the vertex is at (0,0) and the focus is at (0,2) which lies on y-axis, the y-axis is the axis of the parabola. Therefore, equation of the parabola is of the form x^2 = 4ay. thus, we have

x^2 = 4(2)y, i.e., x^2 = 8y.
Q 3155791664

Find the equation of the parabola which is symmetric about the y-axis, and passes through the point (2,–3).

Solution:

Since the parabola is symmetric about y-axis and has its vertex at the origin,
the equation is of the form x^2 = 4ay or x^2 = – 4ay, where the sign depends on whether
the parabola opens upwards or downwards. But the parabola passes through (2,–3)
which lies in the fourth quadrant, it must open downwards. Thus the equation is of
the form x^2 = – 4ay.
Since the parabola passes through ( 2,–3), we have

2^2 = – 4a (–3), i.e., a = 1/3

Therefore, the equation of the parabola is

x^2 = - 4 (1/3)y , i.e 3x2 = – 4y.