Parabola

A parabola is a U-shaped plane curve where any point is at an equal distance from a fixed point (known as the focus) and from a fixed straight line which is known as the directrix. Parabola is an integral part of conic section topic and all its concepts parabola are covered here.

TABLE OF CONTENTS

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What is Parabola?

Section of a right circular cone by a plane parallel to a generator of the cone is a parabola. It is a locus of a point, which moves so that distance from a fixed point (focus) is equal to the distance from a fixed line (directrix)

  • Fixed point is called focus
  • Fixed line is called directrix

Parabola

Standard Equation of Parabola

The simplest equation of a parabola is y2 = x when the directrix is parallel to the y-axis. In general, if the directrix is parallel to the y-axis in the standard equation of a parabola is given as:

y2 = 4ax

If the parabola is sideways i.e., the directrix is parallel to x-axis, the standard equation of a parabole becomes,

x2 = 4ay

Apart from these two, the equation of a parabola can also be y2 = 4ax and x2 = 4ay if the parabola is in the negative quadrants. Thus, the four equations of a parabola are given as:

    1. y2 = 4ax
    2. y2 = – 4ax
    3. x2 = 4ay
    4. x2 = – 4ay

Parabola Equation Derivation

In the above equation, “a” is the distance from the origin to the focus. Below is the derivation for the parabola equation. First, refer to the image given below.

Equation of Parabola

From definition,

SPPM=1\frac{SP}{PM}=1

SP = PM

(xa)2+y2=x+a1\sqrt{{{\left( x-a \right)}^{2}}+{{y}^{2}}}=\left| \frac{x+a}{1} \right|

(x – a)2 + y2 = (x + a)2

y2=4axy^{2}=4axStandard equation of Parabola.

Latus Rectum of Parabola

The latus rectum of a parabola is the chord that passes through the focus and is perpendicular to the axis of the parabola.

Latus Rectum of Parabola

LSL’ Latus Ractum

= 2(4a.a)2\left( \sqrt{4a.a} \right)

= 4a (length of latus Rectum)

Note: – Two parabola are said to be equal if their latus rectum are equal.

Parametric co-ordinates of Parabola

For a parabola, the equation is y2 = -4ax. Now, to represent the co-ordinates of a point on the parabola, the easiest form will be = atand y = 2at as for any value of “t”, the coordinates (at2, 2at) will always satisfy the parabola equation i.e. y2 = 4ax. So,

Any point on the parabola

y2 = 4ax (at2, 2at)

where ‘t’ is a parameter.

Video Lesson on Parabola

Focal Chord and Focal Distance

Focal chord Any chord passes through the focus of the parabola is a fixed chord of the parabola.

Focal Chord of Parabola

Focal Distance: The focal distance of any point p(x, y) on the parabola y2 = 4ax is the distance between point ‘p’ and focus.

Focal Distance of Parabola

PM = a + x

PS = Focal distance = x + a

General Equations of Parabola

Equation of parabola by definition.

SP = PM

(xα)2+(yβ)2=(x+my+n)22+m2{{(x-\alpha )}^{2}}+{{(y-\beta )}^{2}}=\frac{{{(\ell x+my+n)}^{2}}}{{{\ell }^{2}}+{{m}^{2}}}

The general equation of 2nd degree i.e. ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 if

Δ0\Delta \ne 0 h2=ab{{h}^{2}}=ab

Position of a point with respect to parabola

For parabola

Sy24ax=0,p(x1,y1)S\equiv {{y}^{2}}-4ax=0\,\,\,\,\,\,\,\,\,,\,p(x{}_{1},{{y}_{1}})

S1=y124ax1{{S}_{1}}={{y}_{1}}^{2}-4a{{x}_{1}}

S1<0(insidecurve){{S}_{1}}<0\,\,\,\,\,(inside\,curve)

S1=0(oncurve){{S}_{1}}=0\,\,\,\,\,(on\,curve)

S1>0(outsidecurve){{S}_{1}}>0\,\,\,\,(outside\,curve)

Intersection of a straight line with the parabola y2 = 4ax

Straight line y = mx + c

m slope of straight line

(mx + c)2 – 4ax = 0

m2x2 + 2x(mc – 2a) + c2 = 0

Ax2 Bx + c = 0

B2 – 4AC = discriminant D

D = 0

c=a/mc={}^{a}/{}_{m}

D > 0

mc – a > 0: Straight line intersect the curve

D < 0 (mc – a) < 0: Straight line not touching the curve

Tangent to a Parabola

Tangent at point (x1, y1)

y2 = 4ax (parabola)

equation of Tangent

yy1y12=2a(xx1)y{{y}_{1}}-{{y}_{1}}^{2}=2a(x-{{x}_{1}})

yy14ax1=2a(xx1)y{{y}_{1}}-4a{{x}_{1}}=2a(x-{{x}_{1}})

yy1=2a(x+x1)y{{y}_{1}}=2a(x+{{x}_{1}})

⇒ Point (x1y1)({{x}_{1}}\,{{y}_{1}})

Tangent in slope (m) form:

y2 = 4ax

Let equation of Tangent y = mx + c

From the previous illustration

y = mx + c touches curve at a point

so , c = 9/mc\text{ }=~{}^{9}/{}_{m}

equation of Tangent :- y = mx +  a/m~{}^{a}/{}_{m}

so, point of Tangency is (a/m2,2am)\left( {}^{a}/{}_{{{m}^{2}}},\frac{2a}{m} \right)

Tangent in parameter form (at2, 2at)

ty = x + at2 where ‘t’ is

parameter

Pair of Tangents from (x1, y1) external points

Let y2 = 4ax, (parabola)

P(x1, y1) external point then equation of Tangents is given by

SS1 = T2

Sy24ax,S1y124ax1S\equiv {{y}^{2}}-4\,ax,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{S}_{1}}\equiv {{y}_{1}}^{2}-4a{{x}_{1}}

Tyy12a(xx1)T\equiv y{{y}_{1}}-2a(x-{{x}_{1}})

Chord of contact:

Equation of chord of contact of Tangents from a point p(x1, y1) to the parabola y2 = 4ax is given by T = 0

i.e., yy1 – 2a(x + x1) = 0

Equation of QS T = 0

Normal to the parabola:

Normal to the point p(x1, y1) since normal is perpendicular to Tangent so slope of normal be will

1/SlopeofTangent{}^{-1}/{}_{Slope\,of\,Tangent}

slope of normal at ‘p’ (x1 y1) is y12a\frac{-{{y}_{1}}}{2a}

equation of normalyy1=y12a(xx1)y-{{y}_{1}}=\frac{-{{y}_{1}}}{2a}(x-{{x}_{1}})

Normal in term of ‘m’:

(slopeofnormal)m=dxdyy2=4ax\left( slope\,of\,normal \right)\Rightarrow m=-\frac{dx}{dy\,\,}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{y}^{2}}=4ax\,\,

y1=2am{{y}_{1}}=-2am

m=y12ax1=am2\,\,\,\,\,\,\,m=\frac{-{{y}_{1}}}{2a}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}_{1}}=a{{m}^{2}}\,\,\,\,

y=mx2amam3y=mx-2am-a{{m}^{3}}

m=dxdym=\frac{-dx}{dy}

Equation of normal at point (am2, – 2am)

Normal at point (at2, 2at)

T parameter

y = tx + 2at + at3

Important Properties of Focal Chord

      1. If chord joining P=(at12,2at1)P=(at_{1}^{2},2a{{t}_{1}}) and Q=(at22,2at2)Q=(at_{2}^{2},2a{{t}_{2}})is focal chord of parabola y2=4ax{{y}^{2}}=4ax then t1t2=1{{t}_{1}}{{t}_{2}}=-1.
      2. If one extremity of a focal chord is (at12,2at1)(at_{1}^{2},2a{{t}_{1}}) then the other extremity (at12,2at2)(at_{1}^{2},2a{{t}_{2}}) becomes (at12,2at1)\left( \frac{a}{t_{1}^{2}},-\frac{2a}{{{t}_{1}}} \right).
      3. If point P(at2,2at)P(a{{t}^{2}},2at) lies on parabola y2=4ax{{y}^{2}}=4ax, then the length of focal chord PQ is a(t+1/t)2a{{(t+1/t)}^{2}}.
      4. The length of the focal chord which makes an angle θ with positive x-axis is 4acosec2θ4a\cos e{{c}^{2}}\theta.
      5. Semi latus rectum is harmonic mean of SP and SQ, where P and Q are extremities of latus rectum. i.e., 2a=2SP×SQSP+SQor1SP+1SQ=1a2a=\frac{2SP\times SQ}{SP+SQ}\,or\frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a}
      6. Circle described on focal length as diameter touches tangent at vertex.
      7. Circle described on focal chord as diameter touches directrix.

Important Properties of focal chord, Tangent and normal of Parabola

      • 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 directix.

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

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

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

Intersect at y-axis, at u = 0

Four Common Forms of a Parabola:

Form: y2 = 4ax y2 = – 4ax x2 = 4ay x2 = – 4ay
Vertex: (0, 0) (0,0) (0, 0) (0, 0)
Focus: (a, 0) (-a, 0) (0, a) (0, -a)
Equation of the directrix: x = – a x = a y = – a y = a
Equation of the axis: y = 0 y = 0 x = 0 x = 0
Tangent at the vertex: x = 0 x = 0 y = 0 y = 0

Practice Problems on Parabola

Illustration 1: Find the vertex, axis, directrix, tangent at the vertex and the length of the latus rectum of the parabola 2y2+3y4x3=02{{y}^{2}}+3y-4x-3=0.

Solution: The given equation can be re-written as (y+34)2=2(x+3332){{\left( y+\frac{3}{4} \right)}^{2}}=2\left( x+\frac{33}{32} \right)

which is of the form Y2=4aX{{Y}^{2}}=4aXwhere Y=y+34,X=x+3332,4a=2Y=y+\frac{3}{4},\,X=x+\frac{33}{32},\,4a=2.

Hence the vertex is X=0,Y=0X=0,Y=0 i.e. (3332,34)\left( -\frac{33}{32},-\frac{3}{4} \right).

The axis is y+34=0y=34y+\frac{3}{4}=0\Rightarrow y=-\frac{3}{4}.

The directix is X=a=0X=a=0

x+3332+12=0x=4932\Rightarrow x+\frac{33}{32}+\frac{1}{2}=0\Rightarrow x=-\frac{49}{32}

The tangent at the vertex is X=0orx+3332=0x=3332X=0\,or\,x+\frac{33}{32}=0\Rightarrow x=-\frac{33}{32}.

Length of the latus rectum = 4a = 2.

Illustration 2: Find the equation of the parabola whose focus is (3, -4) and directix x – y + 5 = 0.

Solution: Let P(x, y) be any point on the parabola. Then

(x3)3+(y+4)2=xy+51+1\sqrt{{{(x-3)}^{3}}+{{(y+4)}^{2}}}=\frac{\left| x-y+5 \right|}{\sqrt{1+1}}

(x3)2+(y+4)2=(xy+5)22\Rightarrow {{(x-3)}^{2}}+{{(y+4)}^{2}}=\frac{{{(x-y+5)}^{2}}}{2}

x2+y2+2xy22x+26y+25=0\Rightarrow {{x}^{2}}+{{y}^{2}}+2xy-22x+26y+25=0

(x+y)2=22x26y25\Rightarrow {{(x+y)}^{2}}=22x-26y-25

Illustration 3: Find the equation of the parabola having focus (-6, -6) and vertex (-2, 2).

Solution: Let S(6, -6) be the focus and A(-2, 2) the vertex of the parabola. On SA take a point K (x1 , y1) such that SA = AK. Draw KM perpendicular on SK. Then KM is the directix of the parabola.

Since A bisects SK, (6+x12,6+y12)=(2,2)\left( \frac{-6+{{x}_{1}}}{2},\frac{-6+{{y}_{1}}}{2} \right)=(-2,2)

6+x1=4and6+y1=4or(x1,y1)=(2,10).\Rightarrow -6+{{x}_{1}}=-4\,and\,-6+{{y}_{1}}=4\,or\,({{x}_{1}},{{y}_{1}})=(2,10).

Hence the equation of the directrix KM is y – 10 = m(x+2) ……(1)

Also gradient of SK=10(6)2(6)=168=2;m=12SK=\frac{10-(-6)}{2-(-6)}=\frac{16}{8}=2;\,m=\frac{-1}{2}


So that equation (1) becomes

y10=12(x2)y-10=\frac{1}{2}(x-2) or x+2y22=0x+2y-22=0 is the directrix.

Next, let PM be a perpendicular on the directrix KM from any point P(x, y) on the parabola.

From SP = PM, the equation of the parabola is

{(x+6)2+(y+6)2}=x+2y22(12+22)\sqrt{\left\{ {{(x+6)}^{2}}+{{(y+6)}^{2}} \right\}}=\frac{x+2y-22}{\sqrt{({{1}^{2}}+{{2}^{2}})}}

Illustration 4: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2=12x{{y}^{2}}=12x.

Solution: The given equation is y2=12x{{y}^{2}}=12x.

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with y2=4ax{{y}^{2}}=4ax, we get 4a=12a4a=12a or a=3a=3.

Coordinates of the focus are given by (a, 0) i.e., (3, 0).

Since the given equation involves y2, the axis of the parabola is the y-axis.

Equation of directix is x=ax=-a, i.e., x=3x=-3.

Length of latus rectum = 4a = 4 x 3 = 12.

Illustration 5: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2=16y{{x}^{2}}=-16y.

Solution: The given equation is x2=16y{{x}^{2}}=-16y.

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with x2=4ay{{x}^{2}}=-4ay, we get 4a=16-4a=-16 or a=4a=4.

Coordinates of the focus = (0, -a) = (0, -4).

Since the given equation involves x2{{x}^{2}}, the axis of the parabola is the y-axis.

Equation of directrix, y = a i.e. = 4.

Length of latus rectum = 4a = 16.

Illustration 6: If the parabola y2 = 4x and x2 = 32y intersect at (16, 8) at an angle θ, then find the value of θ.

Solution: The slope of the tangent to y2 = 4x at (16, 8) is given by

m1=(dydx)(16,8)=(42y)(16,8)=28=14{m}_{1}={\left( \frac{dy}{dx} \right)}_{(16,8)}={{\left( \frac{4}{2y} \right)}_{(16,8)}}=\frac{2}{8}=\frac{1}{4}

The slope of the tangent to x2 = 32y at (16, 8) is given by

m2=(dydx)(16,8)=(2x32)(16,8)=1{m}_{2}={\left( \frac{dy}{dx} \right)}_{(16,8)} ={{\left( \frac{2x}{32} \right)}_{(16,8)}}=1

Tan  θ=1(1/4)1+(1/4)=35Tan \;\theta =\frac{1-(1/4)}{1+(1/4)}=\frac{3}{5}

θ=tan1(35)\Rightarrow \,\,\,\,\,\theta ={{\tan }^{-1}}\left( \frac{3}{5} \right)

Illustration 7: Find the equation of common tangent of y2 = 4ax and x2 – 4ay.

Solution: Equation of tangent to y2 = 4ax having slope m is y=mx+amy=mx+\frac{a}{m}.

It will touch x2 – 4ay, if x2=4a(mx+am){{x}^{2}}=4a\left( mx+\frac{a}{m} \right) has a equal roots.


Thus, 16a2m2= 16a2mm=116{{a}^{2}}{{m}^{2}}=\text{ }-16\frac{{{a}^{2}}}{m}\,\,\,\Rightarrow \,m=-1

Thus, common tangent is y + x + a = 0.

Illustration 8: Find the equation of normal to the parabola y2 = 4x passing through the point (15, 12).

Solution: Equation of the normal having slope m is

y=mx2mm3y=mx-2m-{{m}^{3}}

If it passes through the point (15, 12) then

12=15m2mm312=15m-2m-{{m}^{3}}

m313m+12=0\Rightarrow \,\,\,\,\,{{m}^{3}}-13m+12=0

(m1)(m3)(m+4)=0\Rightarrow \,\,\,\,\,\left( m-1 \right)\left( m-3 \right)\left( m+4 \right)=0

m=1,3,4\Rightarrow \,\,\,\,\,m=1,\,3,\,-4

Hence, equations of normal are:

y=x3,y=3x33andy+4x=72y=x-3,\,y=3x-33\,and\,y+4x=72

Illustration 9: Find the point on y2 = 8x where line x + y = 6 is a normal.

Solution: Slope m of the normal x + y = 6 is -1 and a = 2

Normal to parabola at point (am2, -2am) is

y=mx2amam3y=mx-2am-a{{m}^{3}}

y=x+4+2at(2,4)\Rightarrow \,\,\,\,\,y=-x+4+2\,at\,(2,4)

x+y=6isnormalat(2,4)\Rightarrow \,\,\,\,\,x+y=6\,is\,normal\,at\,(2,4)

Illustration 10: Tangents are drawn to y2 = 4ax at point where the line lx + my + n = 0 meets this parabola. Find the intersection of these tangents.

Solution: Let the tangents intersects at P (h, k). Then lx + my + n = 0 will be the chord of contact. That means lx + my + n = 0 and yk – 2ax – 2ah = 0 which is chord of contact, will represent the same line.

Comparing the ratios of coefficients, we get

km=2al=2ahn\frac{k}{m}=\frac{-2a}{l}=\frac{-2ah}{n}

h=nl,k=2aml\Rightarrow \,\,\,\,\,h=\frac{n}{l},\,k=-\frac{2am}{l}

Illustration 11: If the chord of contact of tangents from a point P to the parabola If the chord of contact of tangents from a point P to the parabola y2 = 4ax touches the parabola x2=4by, then find the locus of P.

Solution: Chord of contact of parabola y2 = 4ax w.r.t. point P(x1 , y1)

yy1 = 2a(x + x1) ……(1)

This line touches the parabola x2 = 4by.

Solving line (1) with parabola, we have

x2=4b[(2a/y1)(x+x1)]{{x}^{2}}=4b\left[ \left( 2a/{{y}_{1}} \right)\left( x+{{x}_{1}} \right) \right]

or y1x28abx8abx1=0{{y}_{1}}{{x}^{2}}-8abx-8ab{{x}_{1}}=0

According to the question, this equation must have equal roots.

D=0\Rightarrow \,\,\,\,\,D=0\,

64a2b2+32abx1y1=0\Rightarrow \,\,\,\,64{{a}^{2}}{{b}^{2}}+32ab{{x}_{1}}{{y}_{1}}=0

x1y1=2ab\Rightarrow \,\,\,\,\,{{x}_{1}}{{y}_{1}}=-2ab or xy=2abxy=-2ab, which is the required locus.

Real-life applications of a Parabola

1] A liquid is rotated and the forces of gravity in the liquid are responsible for the formation of a parabola. For example when any juice is stirred round along its axis. The level of the juice increases around the edges while it slightly falls in the centre of the glass.

2] The concept of the parabola is used in satellite dishes which helps to reflect the signals and then go to the receiver. Due to parabola’s reflective properties, the signals that go to the satellite will mirror off and back to the receiver soon after reflecting off the focus.

3] The Golden Gate Bridge cables that act as a suspension are parabolas.

4] Water in The Bellagio’s fountains in Las Vegas are exhibited in the shape of parabolas.

5] The reflective properties of parabolas are used in some heaters. The source of heat is at the focus and the heat
is focused in the form of parallel rays.

6] Parabolas are applied in the field of architecture and engineering projects. Parabolic shapes include:
a] The Parabola, a structure in London built in 1962 that consists of a roof made of copper with parabolic and hyperbolic lines.
b] There exists parabolas on each side of its side spans on The Golden Gate Bridge in San Francisco, California.

7] The use of parabolas are extensive when the light needs to be focused. A parabola-shaped reflector helps in focusing light into a beam that can be seen from long distances. It helps in reducing more usage of light and thus improves the surface of the parabola.

8] Solar power industry is aided by the parabolic reflectors to concentrate light.

9] The most frequent example of a parabola is that of the stretched arc of a rocket launch.

10] The parabolic trajectories are in use from decades.

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