Parabola is the chief and easiest topic in the Conic Sections of Co-ordinate Geometry in Mathematics. The name "parabola" is derived from a Latin term that means something similar to "compare" or "balance", and refers to the fact that the distance from the parabola to the focus is always equal to (that is, is always in balance with) the distance from the parabola to the directrix. In practical terms, you'll probably only need iit-jee-coordinate-geometry/parabola.aspx to know that the vertex is exactly midway between the directrix and the focus. Mathematically, a parabola may be defined as:

A parabola is the locus of points in that plane that are equidistant from both the directrix and the focus. Note that the focus does not lie on the directrix. 

The general equation of a parabola opening to the right with vertex at (0, 0), is given by

y2 = 4ax

Geometrical representation of the parabola y2 = 4cx

Also if the vertex is at (x0, y0) instead of (0, 0), the equation of the parabola becomes
(y-y0)2 = 4a (x-x0)
If the parabola instead opens upwards, its equation is

x2 = 4ay

Parabola of the form x2 = 4ay


There are four common forms of parabola according to their axis, with their vertex at origin (0, 0). 

The four common forms of parabola

Another Form: The standard equation of a parabola may also be written as

y = ax2 + bx + c.

But the equation for a parabola can also be written in "vertex form":

y = a(xh)2 + k

In this equation, the vertex of the parabola is the point (h, k).

Important terms:

We discuss some of the basic terminology of parabola:

Focus: The focus of a parabola is a fixed point in the interior of the parabola.

Focal distance: The distance of a point on the parabola from its focus is called the focal distance of the point Focal distance of P = SP = x + a. 

Focal Chord: A chord of the parabola, which passes through its focus, is called Focal chord.  

Vertex: The vertex of a parabola is the point where the parabola crosses its axis. When the coefficient of the x2 term is positive, then the vertex is the lowest point on the graph but in case it is negative the vertex will be the highest point on the graph.

Directrix: A line perpendicular to the axis of symmetry is called the directrix.

Latus Rectum: The latus rectum of a conic section is the chord through a focus parallel to the conic section directrix. The quantity 4a is known as the latus rectum. Half the latus rectum is called the semilatus rectum.


A parabola depicting focus, directrix, vertex and latus rectum.


State the vertex and focus of the parabola having the equation (y – 3)2 = 8(x – 5).


Comparing the given equation with the general equation of parabola and noticing that the h always goes with the x and the k with y, we get the canter at (h, k) = (5, 3).

Now the coefficient of the unsquared part is 4a, here 4a = 8. This gives a=2.

Now notice that ‘a’ is positive and the y part is squared so this is a sideways parabola that opens to the right. The focus is inside the parabola, so it has to be two units to the right of the vertex. Hence, the vertex is given by (5,3) and the focus is at (7,3).

In order to get more clarity, you may refer the video


Find the vertex of the parabola y = 3x2 + 12x – 12


As discussed above the given equation is of the form y = ax2 + bx + c.

So we represent it in the form  y = a(xh)2 + k

Here, a = 3 and b = 12. So, the x-coordinate of the vertex is:

-12/2.3 = -2

Substituting in the original equation to get the y-coordinate, we get:

y = 3(–2)2 + 12(–2) – 12

= –24

So, the vertex of the parabola is at (2, 24).

Finding the end points of latus Rectum 

For finding the end points of latus rectum LL’ of the parabola y2 = 4ax,

we put x = a as the latus rectum passes through focus (a, 0) and therefore we have 

The parabola y2 = 4ax

y2 = 4a2 

⇒ y = + 2a 

Hence the end points are (a, 2a) and (a, – 2a). 

Also LSL’ = |2a – (–2a)| = 4a = length of double ordinate through the focus S.  


Two parabolas are said to be equal when their latus recta are equal.

The important points & lines related to standard Parabola 

1. The points and lines of two parabolas can be interchanged by transformations. 
2. If a > 0 & a < 0 the parabola will be forward opening and backward opening respectively. 
3. If b > 0 & b < 0 the parabola will be forward opening and downward opening respectively. 

The important points & lines related to shifted Parabola 

The forms of the horizontal and vertical parabola having vertex at (h, k) can be obtained by shifting the origin at (h, k) as below


Find the vertex, axis, directrix, tangent at the vertex and the length of the latus rectum of the parabola 2y2 + 3y – 4x – 3 = 0. 


The given equation can be re-written as (y-3/4)2 = 2 (x+33/32) 

Which is of the form Y2 = 4aX. 

Hence the vertex is (-33/32,-3/4). 

The axis is y + 3/4 = 0 and so y = –3/4.

The directrix is X + a = 0. 

 ⇒ x + 33/32+1/2 = 0 ⇒ x = -49/32. 

The tangent at the vertex is x + 33/32 = 0 ⇒ x = – 33/32. 

Length of the latus rectum = 4a = 2.


For which quadratic equation (parabola) is the axis of symmetry x =3? 

Choose the correct one 

y = -x2 + 3x + 5

y = x2 + 6x + 3

y = -x2 + 6x + 2

y = x2 + x + 3


This is basically a twisted question. We know that the axis of symmetry is given by –b/2a. Just check the value of –b/2a in each equation and the one that gives the value as 3 is the required equation. Solving, we get the third equation satisfies the required condition.


The extreme points of the latus rectum of a parabola are (7, 5) and (7, 3). Find the equation of the parabola and the points where it meets the coordinate axes 

Focus of the parabola is the mid-point of the latus rectum. 

⇒ S is (7, 4). Also axis of the parabola is perpendicular to the latus rectum and passes through the focus. Its equation is

y – 4 = 0/(5-3) (x – 7) ⇒ y = 4.

Length of the latus rectum = (5 – 3) = 2.

Hence the vertex of the parabola is at a distance 2/4 = .5 from the focus. We have two parabolas, one concave rightwards and the other concave leftwards. The vertex of the first parabola is (6.5, 4) and its equation is

(y – 4)2 = 2(x – 6.5) and it meets the x-axis at (14.5, 0).

The equation of the second parabola is (y – 4)2 = –2 (x – 7.5).

It meets the x-axis at (–0.5, 0) and the y-axis at (0, 4 + Ö15). 

Parametric Form of a Parabola 

Suppose that the equation of a tangent to the parabola y2 = 4ax. … (i)

is y = mx + c. … (ii)

The abscissa of the points of intersection of (i) and (ii) are given by the equation (mx + c)2 = 4ax. But the condition that the straight line (ii) should touch the parabola is that it should meet the parabola in coincident points hence discriminant should be zero

⇒ (mx – 2a)2 = m2c2 … (iii)

⇒ c = a/m.

Hence, y = mx + a/m is a tangent to the parabola y2 = 4ax, whatever be the value of m.

Equation (mx + c)2 = 4ax now becomes (mx – a/m)2 = 0.

⇒ x = a/m2 and y2 = 4ax

⇒ y = 2a/m.

Thus the point of contact of the tangent y = mx + a/m is (a/m2, 2a/m). 

Taking 1/m = t where t is a parameter, i.e., it varies from point to point. The parabola y2 = 4ax as a parametric form is given by the co-ordinate (at2, 2at) and we refer to it as point ‘t’.


In case of a parabola as given in the figure, area is given by

A = 2/3 base x height

i.e. A = 2/3 x b x h


Prove that the area of the triangle inscribed in the parabola y2 = 4ax is a2 | (t1– t2) (t2 – t3) (t3 – t1)| where t1, t2 and t3 are the vertices.

The three points on the parabola are (at12, 2at1), (at22, 2at2) and (at32, 2at3).

Illustration: Will the graph of the parabola y = -2x2 + 4x - 4 open upward or downward?


Remember that the coefficient of x2 decides whether the parabola will open upward or downward. Hence here since its coefficient is negative so it opens downward.

Parabola is important from the perspective of scoring high in IIT JEE Examination as there are few fixed pattern on which a number of Multiple Choice Questions are framed on this topic. You may also refer the previous year past papers for getting an idea about the type of questions asked.

Propositions on the Parabola 


(i) The tangent at any point P on a parabola bisects the angle between the focal chord through P and the perpendicular from P on the directrix. 

The tangent at P (at2, 2at) is ty = x + at2

It meets the x-axis at T(–at2, 0). 

Hence ST = a (1 + t2). 

Also, SP = √ (a2 (1+t2)2+4a2 t2) = a (1 + t2) = ST, so that 

∠MPT = ∠PTS = ∠SPT ⇒ TP bisects ∠SPM. 

(ii) The portion of a tangent to a parabola cut off between the directrix and the curve subtends a right angle at the focus. 

Let P(at2, 2a), be a point on the parabola y2 = 4ax. 

The tangent at P is ty = x + at2

Point of intersection of the tangent with the directrix x + a = 0 is (–a, at – a/t).

Now, slope of SP is (2at-0)/(at2-a)=2t/(t2-1) 

and slope of SK is (at-a/t-0)/(-a-a)=-(t2-1)/2t 

⇒ (Slope of the SP).(Slope of SK) = –1. 

Hence SP is perpendicular to SK i.e. ∠KSP = 90°. 

(iii) Tangents at the extremities of any focal chord intersect at right angles on the directrix. 

Let P (at2, 2at) and P (at12, 2at1) be the end points of a focal part on the parabola. Then t.t1 = –1. Equations of the tangents at the point P and the point P’ are ty = x + at2 and t1y = x + at12 respectively. 

Let these tangents intersects at a point (h, k). Then h = att1 and k = a(t + t1). 

Since the tangents are perpendicular, tt1 = – 1 ⇒ h – a. 

Hence the locus of the point (h, k) is x = –a which is the equation of the directrix. 

(iv) Any tangent to a parabola and the perpendicular on it from the focus meet on the tangent at the vertex. 

Equation of the perpendicular to the tangent ty = x + at2 … (1) 

From the focus (a, 0) is tx + y = at. … (2) 

and (2) intersect at x = 0 which is the equation of the tangent at the vertex.

Pole and polar of a conic 

The locus of the point of intersection of tangents drawn at the extremities of the chord of the conic drawn through a point is called the polar of that point with respect to the conic. This point itself is called the pole. 

Equation of the polar of a point (x1, y1) with respect to the parabola y2 = 4ax.

Let us draw the chord QR from the point P(x1, y1) and if the tangents drawn from point Q and R meet at the point T(h, k), required locus of (h, k) is polar. Since QR is the chord of contact of tangents from (h, k), it’s equation is 

      ky = 2a(x + h) 

This straight line passes through the point (x1, y1), we have 

      ky1 = 2a(x1 + h) …… (1) 

 Since the relation (1) is true, it follows that point (h, k) always lies on the line. 

      yy1 = 2a(x + x1) …… (2) 

 Hence (2) is the equation to the polar of pole (x1, y1

Co-normal Points: 
The three points on the parabola, the normals at which pass through a common point, are called the co-normal points. 


The locus of the middle point of a system of parallel chords of a parabola is called its diameter.


Let the parabola be 

  y2 = 4ax. …… (i) 

 let y = mx + c …… (ii) 

 be a system of parallel chords to (i) for different chords, only c varies, m remains  constant. 

  y2 = 44a (y – c)/m 

  my2 – 4ay + 4ac = 0 

let y2 and y3 be the roots of (iii), then y2 and y3 are the ordinates of the points where (ii) cuts (i) 

   from (iii), y2 + y3 = 4a/m 

Also, if (x1, y1) be the midpoint of the chord then 

    y1 = (y2-y3)/2 = 2a/m 

∴ Locus of (x1, y1) is y = 2a/m, which is the equation of the diameter. 


y = 2a/m is a straight line parallel to the axis of the parabola.
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