## Parabola

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

y^{2}= 4axAlso if the vertex is at (x

_{0}, y_{0}) instead of (0, 0), the equation of the parabola becomes

(y-y_{0})^{2}= 4a (x-x_{0})

If the parabola instead opens upwards, its equation isx

^{2}= 4ay

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

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

y=ax^{2}+bx+c.But the equation for a parabola can also be written in "vertex form":

y=a(x–h)^{2}+kIn 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: Thevertexof a parabola is the point where the parabola crosses its axis. When the coefficient of thex^{2}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.

Illustration:State the vertex and focus of the parabola having the equation (

y– 3)^{2}= 8(x– 5).

Solution: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

Illustration:Find the vertex of the parabola

y= 3x^{2}+ 12x– 12

Solution:As discussed above the given equation is of the form

y=ax^{2}+bx+c.

So we represent it in the form y=a(x–h)^{2}+kHere,

a= 3 andb= 12. So, thex-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).we put x = a as the latus rectum passes through focus (a, 0) and therefore we have

Finding the end points of latus Rectum

For finding the end points of latus rectum LL’ of the parabola y^{2}= 4ax,y

^{2}= 4a^{2}⇒ 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.

Note:T

woare said to be equal when theirparabolasare equal.latus recta

The important points & lines related to standard Parabola

Note:

1. The points and lines of twoparabolascan be interchanged by transformations.

2. If a > 0 & a < 0 theparabolawill be forward opening and backward opening respectively.

3. If b > 0 & b < 0 theparabolawill 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

**Illustration:**

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

**Solution:**

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

Which is of the form Y^{2} = 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.

**Illustration:**

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

**Choose the correct one**** **

*y *= -*x*^{2} + 3*x *+ 5

*y* = *x*^{2} + 6*x* + 3

*y* = -*x*^{2} + 6*x* + 2

*y* = *x*^{2} + *x* + 3

**Solution: **

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.

**Illustration:**

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

**Solution:**

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** y^{2} = 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} = m^{2}c^{2} … (iii)

⇒ c = a/m.

Hence, y = mx + a/m is a tangent to the **parabola** y^{2} = 4ax, whatever be the value of m.

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

⇒ x = a/m^{2} and y^{2} = 4ax

⇒ y = 2a/m.

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

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

**AREA OF PARABOLA**

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**Illustration: **

Prove that the area of the triangle inscribed in the parabola y^{2} = 4ax is a^{2} | (t_{1}– t_{2}) (t_{2} – t_{3}) (t_{3} – t_{1})| where t1, t2 and t3 are the vertices.

**Solution:**

The three points on the **parabola** are (at_{1}^{2}, 2at_{1}), (at_{2}^{2}, 2at_{2}) and (at_{3}^{2}, 2at_{3}).

**Illustration:** Will the graph of the parabola y = -2*x*^{2} + 4*x* - 4 open upward or downward?

**Solution:**

Remember that the coefficient of *x*^{2 }decides whether the parabola will open upward or downward. Hence here since its coefficient is negative so it opens downward.

**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 (at^{2}, 2at) is ty = x + at^{2}.

It meets the x-axis at T(–at^{2}, 0).

Hence ST = a (1 + t^{2}).

Also, SP = √ (a^{2} (1+t^{2})^{2}+4a^{2} t^{2}) = a (1 + t^{2}) = 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(at^{2}, 2a), be a point on the parabola y^{2} = 4ax.

The tangent at P is ty = x + at^{2}.

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

Now, slope of SP is (2at-0)/(at^{2}-a)=2t/(t^{2}-1)

and slope of SK is (at-a/t-0)/(-a-a)=-(t^{2}-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 (at^{2}, 2at) and P (at_{1}2, 2at_{1}) be the end points of a focal part on the parabola. Then t.t_{1} = –1. Equations of the tangents at the point P and the point P’ are ty = x + at^{2} and t_{1}y = x + at_{1}2 respectively.

Let these tangents intersects at a point (h, k). Then h = att_{1} and k = a(t + t_{1}).

Since the tangents are perpendicular, tt_{1} = – 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 + at^{2} … (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 (x_{1}, y_{1}) with respect to the parabola y^{2} = 4ax.

Let us draw the chord QR from the point P(x_{1}, y_{1}) 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 (x_{1}, y_{1}), we have

ky_{1} = 2a(x_{1} + h) …… (1)

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

yy_{1} = 2a(x + x_{1}) …… (2)

Hence (2) is the equation to the polar of pole (x_{1}, y_{1})

**Co-normal Points: **

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

**Diameter: **

**parabola**is called its diameter.

Let the **parabola** be

y^{2} = 4ax. …… (i)

let y = mx + c …… (ii)

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

y_{2} = 44a (y – c)/m

my_{2} – 4ay + 4ac = 0

let y_{2} and y_{3} be the roots of (iii), then y_{2} and y_{3} are the ordinates of the points where (ii) cuts (i)

from (iii), y_{2} + y_{3} = 4a/m

Also, if (x_{1}, y_{1}) be the midpoint of the chord then

y_{1} = (y_{2}-y_{3})/2 = 2a/m

∴ Locus of (x_{1}, y_{1}) is y = 2a/m, which is the equation of the diameter.

** **

**Note:**

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

**parabola**.