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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 the conic section topic, and all its concepts are covered here.

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Table of Content 

What Is a Parabola?

The 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 the distance from a fixed point (focus) is equal to the distance from a fixed line (directrix).

  • A fixed point is called the focus
  • A fixed line is called the directrix

Parabola

 

Standard Equation of Parabola

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 the x-axis, the standard equation of a parabola 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.

Parabola 1

Since the eccentricity of the parabola is 1, PM /PF = 1

PM = PF

Using the distance formula,

\(\begin{array}{l}\sqrt{{{\left( x-a \right)}^{2}}+{{y}^{2}}}=\left| \frac{x+a}{1} \right|\end{array} \)

(x – a)2 + y2 = (x + a)2

\(\begin{array}{l}y^{2}=4ax\end{array} \)
⇒ Standard 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.

Parabola Latus rectum

LSL’ = √[(a-a)2 + (2a+2a)2]

= √(16a2 )

= 4a (Length of latus rectum)

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

Parametric Co-ordinates of Parabola

Consider a parabola, y2 = 4ax. Now, to represent the coordinates of a point on the parabola, the easiest form will be x = at2 and y = 2at as for any value of “t”, the coordinates (at2, 2at) will always satisfy the parabola equation, i.e., y2 = 4ax.

So, x = at2 and y = 2at are called the parametric equations of the parabola y2 = 4ax (where t is the parameter).

Parametric equations of the parabola x2 = 4ay are x = 2at, y = at2

Video Lesson on Parabola

Definition, Equation and Terminologies of a Parabola

Conormal Points of a Parabola

Properties of Tangents to a Parabola

Properties of Tangents to a Parabola

Properties of Tangents to a Parabola

Position of a Point with Respect to Parabola

Position of a Point with respect to a Parabola

Focal Chord and Focal Distance

Focal chord:  Any chord that passes through the focus of the parabola is the focal chord of the parabola. It cuts the parabola at two distinct points.

Focal Chord of parabola

Focal Distance: The distance of any point, p(x, y) on the parabola from the focus, is the focal distance. PS is the focal distance in the above figure.

Parabola Focal distance

General Equations of Parabola

The general equation of a parabola is given by y = a(x – h)2 + k or x = a(y – k)2 +h. Here, (h, k) denotes the vertex.

y = a(x – h)2 + k is the regular form

x = a(y – k)2 +h is the sidewise form

Position of a point with respect to the parabola

For parabola y2= 4ax, the position of the point P(x1, y1) depends on the following conditions.

S1 = y1 2– 4ax1

If S1>0, point P lies outside the parabola.

If S1< 0, point P lies inside the parabola.

If S1 = 0, point P lies on the parabola.

The intersection of a straight line with the parabola

Let y2 = 4ax be the parabola and y = mx + c is the straight line.

A line can meet a parabola at most two points.

(mx + c)2 – 4ax = 0

m2x2 + x(2mc – 4a) + c2 = 0.

Here discriminant, D = (2mc – 4a)2 – 4m2c2

If the discriminant is zero, then the line is tangent to the parabola.

If the discriminant is less than zero, the line does not meet the parabola at all.

If the discriminant is greater than zero, the line meets the parabola at two distinct points.

D = 0: Straight line is the tangent to the parabola.

tangent to the parabola 1

D > 0: Straight line intersect the parabola at two points.

Line intersect parabola

D < 0: Straight line does not touch the parabola.

Line not touching parabola

Equation of Tangent to a Parabola

A line which touches the parabola exactly at one point is called the tangent to a parabola.

Point Form

The equation of tangent to parabola y2 = 4ax at (x1, y1) is yy1 = 2a(x+x1).

The point of contact of the tangent is (x1, y1).

Parametric Form

The equation of the tangent to the parabola y2 = 4ax at (at2, 2at) is ty = x + at2.

The point of contact of tangent is (at2, 2at)

Slope Form

If m is the slope of the tangent to the parabola y2 = 4ax, then the equation of tangent is y = mx + a/m.

The point of contact of tangent is (a/m2, 2a/m)

Pair of tangents from (x1, y1) external points

Let y2 = 4ax be the parabola and P(x1, y1) be the external point, then the equation of tangents is given by

SS1 = T2

S = y2 – 4ax, S1 = y12 – 4ax1

T = yy1 – 2a(x + x1).

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

chord of contact

The equation of QS is T = 0

Normal to the Parabola

The line perpendicular to the tangent of the parabola at the point of contact is the normal of a parabola.

The equation of normal to a parabola can be given in point form, parametric form and slope form.

Point Form

The equation of normal to the parabola y2 = 4ax at (x1, y1) is given by y – y1 = (-y1/2a)(x – x1).

Slope Form
The equation of normal to the parabola y2 = 4ax at (am2, -2am) is given by y = mx – 2am – am3.

The point of contact is (am2, -2am).

Parametric Form

The equation of normal to the parabola y2 = 4ax at point (at2, 2at) is given by y = -tx + 2at + at3.

Important Properties of Focal Chord

  1. If chord joining P = (at12, 2at1) and Q = (at22, 2at2) is focal chord of parabola y2 = 4ax, then t1t2 = -1.
  2. If one extremity of a focal chord is (at12, 2at1), then the other extremity (at22, 2at2) becomes (a/t12 , -2a/t1).
  3. If the point (at2, 2at) lies on parabola y2 = 4ax, then the length of focal chord PQ is a(t + 1/t)2.
  4. The length of the focal chord, which makes an angle θ with a positive x-axis, is 4a cosec2 θ.
  5. Semi latus rectum is a harmonic mean between the segments of any focal chord.
  6. Circle described on focal length as diameter touches tangent at the vertex. The circle, described on any focal chord of a parabola as diameter, touches the 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 directrix.

tangent at any point P on a parabola

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

tangent at any point P on a parabola 2

  • Tangents at the extremities of any focal chord intersect at right angles on the directrix.

tangent at any point P on a parabola 3

  • The perpendicular drawn from a focus upon any tangent of a parabola lies on the tangent at the vertex.

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
Length of the latus rectum 4a 4a 4a 4a

Equations of Normal to a Parabola

Tangent to a Parabola

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

\(\begin{array}{l}2{{y}^{2}}+3y-4x-3=0\end{array} \)
.

Solution: The given equation can be re-written as

\(\begin{array}{l}{{\left( y+\frac{3}{4} \right)}^{2}}=2\left( x+\frac{33}{32} \right)\end{array} \)

which is of the form

\(\begin{array}{l}{{Y}^{2}}=4aX\end{array} \)
where
\(\begin{array}{l}Y=y+\frac{3}{4},\,X=x+\frac{33}{32},\,4a=2\end{array} \)
.

Hence, the vertex is X = 0, Y = 0, i.e. (-33/32, -3/4).

The axis is

\(\begin{array}{l}y+\frac{3}{4}=0\Rightarrow y=-\frac{3}{4}\end{array} \)
.

The directrix is X = a = 0.

\(\begin{array}{l}\Rightarrow x+\frac{33}{32}+\frac{1}{2}=0\Rightarrow x=-\frac{49}{32}\end{array} \)

The tangent at the vertex is

\(\begin{array}{l}X=0\,or\,x+\frac{33}{32}=0\Rightarrow x=-\frac{33}{32}\end{array} \)
.

Length of the latus rectum = 4a = 2.

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

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

\(\begin{array}{l}\sqrt{{{(x-3)}^{3}}+{{(y+4)}^{2}}}=\frac{\left| x-y+5 \right|}{\sqrt{1+1}}\end{array} \)

\(\begin{array}{l}\Rightarrow {{(x-3)}^{2}}+{{(y+4)}^{2}}=\frac{{{(x-y+5)}^{2}}}{2}\end{array} \)

\(\begin{array}{l}\Rightarrow {{x}^{2}}+{{y}^{2}}+2xy-22x+26y+25=0\end{array} \)

\(\begin{array}{l}\Rightarrow {{(x+y)}^{2}}=22x-26y-25\end{array} \)

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 perpendicularly on SK. Then, KM is the directrix of the parabola.

Since A bisects SK,

\(\begin{array}{l}\left( \frac{-6+{{x}_{1}}}{2},\frac{-6+{{y}_{1}}}{2} \right)=(-2,2)\end{array} \)

\(\begin{array}{l}\Rightarrow -6+{{x}_{1}}=-4\,and\,-6+{{y}_{1}}=4\,or\,({{x}_{1}},{{y}_{1}})=(2,10).\end{array} \)

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

Also gradient of

\(\begin{array}{l}SK=\frac{10-(-6)}{2-(-6)}=\frac{16}{8}=2;\,m=\frac{-1}{2}\end{array} \)

Parabola Ex 3

So, equation (1) becomes

\(\begin{array}{l}y-10=\frac{1}{2}(x-2)\end{array} \)
or
\(\begin{array}{l}x+2y-22=0\end{array} \)
is the directrix.

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

From SP = PM, the equation of the parabola is

\(\begin{array}{l}\sqrt{\left\{ {{(x+6)}^{2}}+{{(y+6)}^{2}} \right\}}=\frac{x+2y-22}{\sqrt{({{1}^{2}}+{{2}^{2}})}}\end{array} \)

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

Solution: The given equation is y2 = 12x.

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

On comparing this equation with y2 = 4ax, we get 4a = 12 or a = 3.

The 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.

The equation of directrix is x = -1, i.e., x = -3

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

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

Solution: The given equation is x2 = -16y.

Here, the coefficient of y is negative.

Hence, the parabola opens downwards.

On comparing this equation with x2 = -4ay, we get,

-4a = -16

a = 4

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

Since the given equation involves x2, 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

\(\begin{array}{l}{m}_{1}={\left( \frac{dy}{dx} \right)}_{(16,8)}={{\left( \frac{4}{2y} \right)}_{(16,8)}}=\frac{2}{8}=\frac{1}{4}\end{array} \)

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

\(\begin{array}{l}{m}_{2}={\left( \frac{dy}{dx} \right)}_{(16,8)} ={{\left( \frac{2x}{32} \right)}_{(16,8)}}=1\end{array} \)

\(\begin{array}{l}\therefore Tan \;\theta =\frac{1-(1/4)}{1+(1/4)}=\frac{3}{5}\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,\,\,\theta ={{\tan }^{-1}}\left( \frac{3}{5} \right)\end{array} \)

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

Solution: Equation of tangent to y2 = 4ax having slope m is

\(\begin{array}{l}y=mx+\frac{a}{m}\end{array} \)
.

It will touch x2 – 4ay, if

\(\begin{array}{l}{{x}^{2}}=4a\left( mx+\frac{a}{m} \right)\ \text{has equal roots.}\end{array} \)

Parabola Ex 7

Thus,

\(\begin{array}{l}16{{a}^{2}}{{m}^{2}}=\text{ }-16\frac{{{a}^{2}}}{m}\,\,\,\Rightarrow \,m=-1\end{array} \)

Thus, the 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: The equation of the normal having slope m is

\(\begin{array}{l}y=mx-2m-{{m}^{3}}\end{array} \)

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

\(\begin{array}{l}12=15m-2m-{{m}^{3}}\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,\,\,{{m}^{3}}-13m+12=0\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,\,\,\left( m-1 \right)\left( m-3 \right)\left( m+4 \right)=0\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,\,\,m=1,\,3,\,-4\end{array} \)

Hence, the equations of normal are:

\(\begin{array}{l}y=x-3,\,y=3x-33\,and\,y+4x=72\end{array} \)

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

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

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

\(\begin{array}{l}y=mx-2am-a{{m}^{3}}\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,\,\,y=-x+4+2\,at\,(2,4)\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,\,\,x+y=6\,is\,normal\,at\,(2,4)\end{array} \)

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

Solution: Let the tangents intersect 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 a chord of contact, will represent the same line.

Comparing the ratios of coefficients, we get

\(\begin{array}{l}\frac{k}{m}=\frac{-2a}{l}=\frac{-2ah}{n}\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,\,\,h=\frac{n}{l},\,k=-\frac{2am}{l}\end{array} \)

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

\(\begin{array}{l}{{x}^{2}}=4b\left[ \left( 2a/{{y}_{1}} \right)\left( x+{{x}_{1}} \right) \right]\end{array} \)

or

\(\begin{array}{l}{{y}_{1}}{{x}^{2}}-8abx-8ab{{x}_{1}}=0\end{array} \)

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

\(\begin{array}{l}\Rightarrow \,\,\,\,\,D=0\,\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,\,64{{a}^{2}}{{b}^{2}}+32ab{{x}_{1}}{{y}_{1}}=0\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,\,\,{{x}_{1}}{{y}_{1}}=-2ab\end{array} \)
or
\(\begin{array}{l}xy=-2ab\end{array} \)
, which is the required locus.

Illustration 12:

\(\begin{array}{l}\text { Length of intercept by the line 4y = 3x – 48 on the parabola} \ y^{2}=64 x \ is\\ (1) \frac{9}{1600}\\ (2) \frac{1600}{9}\\ (3) \frac{160}{9}\\ (4) \text { None of these}\\ Solution:\\ \begin{array}{l} 4 y=3 x-48 \Rightarrow m=3 / 4, c=-12 \\ y^{2}=64 x \Rightarrow a=16 \end{array}\\ \begin{aligned} \text { Length of intercept } &=\frac{4}{m^{2}} \sqrt{a\left(1+m^{2}\right)(a-m c)} \\ &=\frac{4}{9} \times 16 \sqrt{16\left(1+\frac{9}{16}\right)\left(16+12 \times \frac{3}{4}\right)} \\ &=\frac{1600}{9} \end{aligned}\\ Answer: [2]\end{array} \)

Illustration 13:

\(\begin{array}{l}\text \ If \ two \ normals \ are \ drawn \ from \ any \ point \ to \ the \ parabola \ y^{2}=4 a x \ makes \ angle \ \alpha \ and \ \beta \ with \ the \ axis \ such \ that \ \tan \alpha \cdot \tan \beta=2, \ then \ locus \ of \ this \ point \ is\\
(1) y^{2}=4 a x\\
(2) x^{2}=4 a y\\
(3) y^{2}=-4 a x\\
(4) x^{2}=-4 ay\\
Solution: \\
\text \ Let \ the \ point \ is \ (\mathrm{h}, \mathrm{k}). \\
\text { The equation of any normal to the parabola } \mathrm{y}^{2}=4 \mathrm{ax} \ is\\
y=m x-2 a m-a m^{3}\\
\text {Passes through ( h, k)}\\
\begin{array}{l}
k=m k-2 a m-a m^{3} \\
a m^{3}+m(2 a-h)+k=0
\end{array}\\
\mathrm{m}_{1}, \mathrm{~m}_{2}, \mathrm{~m}_{3} \text { are roots of the equation}\\
m_{1} m_{2} m_{3}=-\frac{k}{a}\\
m_{1} m_{2}=2, m_{3}=-\frac{k}{2 a}\\
\mathrm{m}_{3} \text \ is \ root \ of \ (i)\\
a\left(-\frac{k}{2 a}\right)^{3}-\frac{k}{2 a}(2 a-h)+k=0 \Rightarrow k^{2}=4 a h\\
\text {Thus locus is }y^{2}=4 a x\\
Answer: [1]\end{array} \)

Illustration 14:

\(\begin{array}{l}\text \ If \ a \ normal \ chord \ of \ the \ parabola \ y^{2}=4 a x \ subtend \ a \ right \ angle \ at \ the \ vertex, \ its \ slope \ is\\ (1) \pm 1\\ (2) \pm \sqrt{2}\\ (3) \pm \sqrt{3}\\ (4) \text {None of these}\\ Solution: If P\left(a t_{1}^{2}, 2 a t_{1}\right) \text {be one end of the normal, the other say } Q\left(a t_{2}^{2}, 2 a t_{2}\right) then t_{2}=-t_{1}-\frac{2}{t_{1}}\\ \text {Again slope of OP }=\frac{2 \mathrm{at}_{1}}{\mathrm{at}_{1}^{2}}=\frac{2}{\mathrm{t}_{1}}\\ \text {Slope of OQ}=\frac{2}{t_{2}} \quad \therefore \frac{2}{t_{1}} \times \frac{2}{t_{2}}=-1\\ \Rightarrow t_{1} t_{2}=-4\\ \text {From (1) and (2)}\\ -\frac{4}{t_{1}}=-t_{1}-\frac{2}{t_{1}} \Rightarrow \frac{2}{t_{1}}=t_{1} \Rightarrow t_{1}^{2}=2 \Rightarrow t_{1}=\pm \sqrt{2}\\ Answer: [2]\end{array} \)

Parabola Rapid Revision for JEE

Parabola Solved Questions

Parabola Solved Questions

Parabola – Top 12 Most Important and Expected JEE Questions

Frequently Asked Questions

Q1

What do you mean by the conjugate axis of a parabola?

The conjugate axis is the line perpendicular to the transverse axis of the parabola and passes through the vertex of the parabola.

Q2

What is the eccentricity of a parabola?

The eccentricity of a parabola equals 1.

Q3

Give the equation of the chord of contact of tangents from a point P(x1, y1) to the parabola y2 = 4ax.

The equation of the chord of contact of tangents from a point P(x1, y1) to the parabola y2 = 4ax is given by yy1 – 2a(x + x1) = 0.

Q4

List two applications of a parabola.

Parabolic arch is used in architecture.
Parabolic reflectors are used in satellite dishes, reflecting telescopes, etc.

Q5

What do you mean by the foci of a parabola?

The focus of a parabola lies on the axis of the parabola. For the parabola having equation y2 = 4ax and having the x-axis as its axis, the focus is given by F (a, 0).

Q6

What do you mean by the vertex of a parabola?

The vertex of a parabola is the point at the intersection of the parabola and the line of symmetry. It is the point where the parabola makes its sharpest turn. It will denote the maximum and minimum points of the parabola. For a parabola y2 = 4ax, the vertex is (0, 0).

Q7

How is the graph of a parabola shaped?

The graph of a parabola is a U-shaped curve.

Q8

Give the formula for the length of the latus rectum of a parabola.

If y2 = 4ax denotes the parabola, then the length of the latus rectum is given by 4a.

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  1.  rahul
    August 6, 2019 at 7:11 pm

    good and best to get suceed in a chapter

    Reply