Parabola | Standard Equations | Latus Ractum, Focus, directrix, Focal Chord & Focal Distance, Tangent, Normal, Solved Examples

Class 11 Mathematics – Written by Neeraj Anand
Published by ANAND TECHNICAL PUBLISHERS

Introduction to Parabola

A parabola is a conic section formed when a plane cuts a right circular cone parallel to its slant height. It is a U-shaped curve where every point is equidistant from a fixed point (called the focus) and a fixed line (called the directrix).

Equation of Parabola

Equation of Parabola can vary depending on its orientation and the position of its vertex, but one common form is:

y = ax² + bx + c

Here, a, b, and c are constants. The shape of the parabola depends primarily on the value of a:

• If a > 0, the parabola opens upwards.
• If a < 0, the parabola opens downwards.

Standard Equations of Parabola

Standard Equation of Parabola is given as follows:

y² = 4ax

In this form, directrix is parallel to the y-axis.

If directrix is parallel to the x-axis, then the standard equation of a parabola is given by,

x² = 4ay

If the parabolas are drawn in alternate quadrants, then their equation is given as y² = -4ax and x² = -4ay. 

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 Axis	y = 0	y = 0	x = 0	x = 0
Equation of Directrix	x = – a	x = a	y = – a	y = a
Length of Latus Rectum	4a	4a	4a	4a

Vertex Form of a Parabola

General equation of a parabola is given by

y = a(x – h)² + k

or x = a(y – k)² +h where (h, k) denotes the vertex.

(Regular form) y = a(x – h)² + k

(Sidewise from) x = a(y – k)² + h

Parametric Coordinates of a Parabola

For a parabola, y² = 4ax, if we take x = at² and y = 2at for any value of “t” they will satisfy the equation of a parabola, the coordinates (at², 2at) is termed as parametric coordinate, and “t” is called as the parameter.

Thus, x = at² and y = 2at are called the parametric equations of the parabola y² = 4ax

Similarly, parametric form of the parabola x² = 4ay are x = 2at, y = at²

Equation of Tangent to a Parabola

Tangents are lines that touch the curve only at a single point. So a line that touches the parabola exactly at one single point is called the tangent to a parabola.

Equation of Tangent in Point Form

For the given parabola y² = 4ax equation of the tangent at point (x₁, y₁) is given by:

yy₁ = 2a(x+x₁)

where, 

(x₁, y₁) is the point of contact between the tangent and the curve.

Equation of Tangent in Parametric Form

For the given parabola y² = 4ax equation of the tangent at point (at², 2at) is given by:

ty = x + at² 

where, 

(at², 2at) is the point of contact between the tangent and the curve.

Equation of Tangent in Slope Form

For the given parabola y² = 4ax with slope m equation of the tangent at point (a/m², 2a/m) is given by

y = mx + a/m 

where,

(a/m², 2a/m) is the point of contact between the tangent and the curve.

Pair of Tangent from an External Point

Pair of tangents from an external point to any conic is given by SS₁ = T² where for parabola y² = 4ax, S = y² – 4ax, S₁ = y₁² -4ax₁ and T =  yy₁ – 2a(x + x₁).

Thus, the equation of pair of tangents from an external point becomes:

(y² – 4ax)( y₁² -4ax₁) = [yy₁ – 2a(x + x₁)]²

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.

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

y²=4ax

⇒ 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.

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

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.

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 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.

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.

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

D < 0: Straight line does not touch the 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

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.

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

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

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

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

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

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

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

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:

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

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.

Summary

Standard Equations of a Parabola

The general equation of a parabola depends on its orientation and the position of its vertex, focus, and directrix. The four standard forms of a parabola with the vertex at the origin (0,0) are:

1. Parabola Opening Right: The focus lies on the positive x-axis.
2. Parabola Opening Left: The focus lies on the negative x-axis.
3. Parabola Opening Upward: The focus lies on the positive y-axis.
4. Parabola Opening Downward: The focus lies on the negative y-axis.

Focus, Directrix, and Axis of Symmetry

• The focus of a parabola is the point where all reflected rays parallel to the axis converge.
• The directrix is a fixed line perpendicular to the axis of symmetry, maintaining equal distance from any point on the parabola.
• The axis of symmetry is the vertical or horizontal line passing through the focus and vertex, dividing the parabola into two symmetric halves.

Latus Rectum

The latus rectum is a line segment perpendicular to the axis of symmetry that passes through the focus. The length of the latus rectum is an important property and depends on the focal length of the parabola.

Focal Chord & Focal Distance

• A focal chord is any line segment passing through the focus and lying on the parabola.
• The focal distance of any point on the parabola is the distance from that point to the focus.

Tangent & Normal to a Parabola

• A tangent to a parabola is a straight line that touches the curve at only one point. It helps in solving problems related to reflection properties.
• A normal to a parabola is a line perpendicular to the tangent at the point of contact.

Applications of Parabola

Parabolas have various real-world applications, including:

• Physics: Used in projectile motion equations.
• Engineering: Parabolic reflectors in satellite dishes and telescopes.
• Architecture: Parabolic arches in bridges and domes.

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