PARABOLA - Safe Hands, Akola

Notes & Example on Parabola

Date: March 28, 2020 By: Ashok Kumar

PARABOLA A conic section is a section cut off from a right circular cone by a plane in various ways. The shape of the section depends upon the position of the cutting plane.

1. If the plane passes through the axis of the cone, the curve of intersection will be a

pair of straight lines.

2. If the plane is perpendicular to the axis of the cone, the curve of intersection will be a circle.

3. If the plane is parallel to a generator of the cone (Say, PQ), the curve of intersection

will be a parabola.

V

AP Q

B

O

Plane

P QO

V

Generator

Parabola

Plane

P QO

V

Ellipse

Plane

P QO

V

Plane

Circle

Circular

base

4. If the plane cuts the axis of the cone at an angle (0 < < /2), the curve of intersection will be an ellipse.

5. If the plane is parallel to the axis of the cone, the curve of intersection will be a

hyperbola.

1.1 DEFINITION OF CONIC A conic section or conic is the locus of a point, which moves such that its distance from a fixed point is in a constant ratio to its distance from a fixed straight line, not passing through the fixed point.

(i) The fixed point is called the focus.

(ii) The fixed straight line is called the directrix.

(iii) The constant ratio is called the eccentricity of the conic and is denoted by e.

(iv) When the eccentricity is unity; i.e., e = 1, the conic is called a parabola; when e < 1, the conic is called an ellipse; and when e > 1, the conic is called a hyperbola.

(v) The line of symmetry of the conic section is called its axis.

(vi) A point of intersection of a conic with its axis is called vertex.

(vii) The chord of a conic which passes through the focus and perpendicular to the axis is called the latus rectum.

1.2 GENERAL EQUATION OF CONIC

Let S (, ) be the focus and Ax + By + C = 0 be the equation of the directrix QN of the conic section.

Let P (x, y) be any point on it and let PN QN. If ‘e’ be the eccentricity of the conic, then by definition,

PN

PS = e PS2 = e2 . PN2 ……(1)

(x – )2 + (y – )2 = e2

2

22 BA

CByAx

which on simplification takes the form

Q

N P (x, y)

S (, )

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ……(2)

Hyperbola

Axis

where a, b, c, f, g and h are constants. (2) being the locus of P, is the equation of the conic.

1.3 RECOGNITION OF CONICS The general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents

A pair of straight line If = 0 where = abc + 2fgh - af2 - bg2 - ch2 , 2h ab

A circle if 0, h = 0, a = b

A parabola if 0, ab - h2 = 0

An ellipse if 0, ab - h2 > 0

An hyperbola if 0, ab – h2 < 0

Example -1: Find the equation of the parabola whose focus is (3, -4) and directrix is x - y + 5 = 0.

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

11

5yx4y3x

22

(x - 3)2 + (y + 4)2 =

2

5yx2

x2 + y2 + 2xy - 22x + 26y + 25 = 0 (x + y)2 = 22x - 26y – 25.

Example -2: If (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then its equation is

(A) x2 + 8y = 32 (B) y2 + 8x = 32 (C) x2 - 8y = 32 (D) y2 - 8x = 32

Solution: AS = 2 = a. Vertex (0, 4) lies on y-axis. Hence the parabola X2 = -4aY is a downward parabola as focus is below the vertex.

Or (x –0)2 = -4 2(y –4) Or x2 + 8y = 32 Hence (A) is the correct answer.

A (0, 4)

S (0, 2)

X

Y’

Example -3: Co-ordinates of the focus of the parabola x2 –4x –8y –4 = 0 are (A) (0, 2) (B) (2, 1) (C) (1, 2) (D) (-2, -1)

Solution: (x –2)2 = 8(y + 1) Focus:- X = 0

x –2 = 0 x = 2

Similarly Y = a y + 1 = 2 y = 1

Focus is (2, 1) Hence (B) is the correct answer.

Example -4: If the point (2, 3) is the focus and x = 2y + 6 is the directrix of a parabola, find

(I) The equation of the axis (ii) The co-ordinates of the vertex

(iii) Length of the latus rectum (iv) Equation of the latus rectum

Solution: (i) We know that the axis of a parabola is the line through the focus and

perpendicular to the directrix.

The equation of any line passing through the focus (2, 3) is

y – 3 = m(x-2) y – mx = 3 –2m.

If the line be perpendicular to the directrix x – 2y = 6 we have, m + 2=0 m = –2

Hence the equation of the axis is y – 3 = –2(x-2) 2x + y = 7

(ii) The co-ordinates of the point of intersection (say) A of the directrix

x – 2y = 6 and the axis 2x + y = 7 are obtained by solving the two equations; thus

they are (4, -1). Since the vertex O is the middle point of A (4, –1) and the focus S (2,

3); the co-ordinates of the vertex are ,2

13,

2

24

i.e. (3, 1).

(iii) Since OS = 5)31()23( 22 , the length of the latus rectum

= 4OS = 54 .

(iv) Since the latus rectum is the line through the focus parallel to the directrix, its

equation is x-2y+c=0, where c is given by 2 – 2.3 + c = 0, i.e. c = 4

1.4 STANDARD EQUATION OF A PARABOLA:

Let S be the focus, ZM the directrix and P the moving point.

Draw SZ perpendicular from S on the directrix. Then SZ is

the axis of the parabola. Now the middle point of SZ, say A,

will lie on the locus of P, i.e., AS = AZ. Take A as the origin,

the x-axis along AS, and the y- axis along the perpendicular

to AS at A, as in the figure.

Let AS = a, so that ZA is also a. Let (x, y) be the coordinates

of the moving point P. Then MP = ZN = ZA + AN = a + x.

But by definition MP = PS MP2 = PS2

So that, (a + x)2 = (x – a)2 + y2.

Hence, the equation of parabola is y2 = 4ax.

Z

M

S X

N

P

A

Y

2.1 DIFFERENT FORMS OF THE PARABOLA Form y2 = 4ax Y2 = –4ax x2 = 4ay x2 = –4ay

Shape of the parabola

X

Y

S

L

L

A

N

P(x,y)

(0,0)

(a,0)

x a

2y 4ax

x

x = a

A

(0, 0)

P (x, y)

Y

2y 4ax

x

y = -b

S (0, a)L' L

A

(0, 0)

P (x, y)

Y

2x 4ay

x

y =a

S (0, -a)L' L

A(0, 0)

P (x, y)

Y2x 4ay

Vertex (0, 0) (0, 0) (0, 0) (0, 0)

Focus (a, 0) (–a, 0) (0, a) (0, –a)

Equation of Directrix

x = –a x = a y = –a y = a

Equation of axis y = 0 y = 0 x = 0 x = 0

Tangent at the vertex

x = 0 x = 0 y = 0 y = 0

Parametric Equation

x = at2, y = 2at x = –at2, y = 2at x = 2at, y = at2 x = 2at, y = –at2

Focal Distance of (x1, y1)

x1 + a –x1 + a y1 + a –y1 + a

Equation of Tangent

(a) At the point (x1, y1)

yy1 = 2a(x + x1) yy1 = –2a(x + x1) xx1 = 2a(y + y1) xx1 = –2a(y + y1)

(b) in terms of m y = mx + a/m y = mx – a/m x = my + a/m x = my – a/m

(c) in terms of t yt = x + at2 yt = –x + at2 xt = y + at2 xt = –y + at2

Equation of Normal

(a) At the point (x1, y1)

y – y1 = a2

y1(x – x1) y – y1 =

a2

y1(x – x1) x – x1 =

a2

x 1(y – y1) x – x1 =

a2

x1(y – y1)

(b) in terms of m y = mx – 2am – am3 y = mx +2am + am3 x = my –2am – am3 x = my +2am + am3

(c) in terms of t y=–tx+2at+at3 y = tx + 2at + at3 x = –ty + 2at + at3 x = ty + 2at + at3

Latus Rectum of the Parabola:

Let the given parabola be y2 = 4ax.

In the figure LSL is the latus rectum.

Also by definition, LSL =2 a4a.a4 = double ordinate

through the focus S.

y2 = 4ax

L

S(a, 0)

L

xA

y

Notes:

Any chord of the parabola y2 = 4ax which is perpendicular to its axis is called the double

ordinate.

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

Example 5: Find the equation of parabola whose focus is (1, - 1), and whose vertex is (2, 1). Also find equation of its axis and latus rectum.

Solution: As we know that vertex is mid point of focus and point of intersection of

directrix with the axis so point P(h, k) is 3h22

1h

11 3

2

kk

P is (3, 3)

slope of axis = 212

)1(1

equation of axis = y – 1 = 2 (x – 2) 2x – y = 3

P O S

equation of directrix y – 3 = -2

1(x – 3) x + 2y = 9

equation of parabola is (x – 1)2 + (y + 1)2 = 5

)9y2x( 2

equation of latus roctum

y + 1 = 2

1(x – 1) x + 2y + 1 = 0

Parametric Equations of a Parabola:

If the coordinates of any point (x, y) on a curve can be expressed as functions of a variable t, given

by x = (t), y = (t) ……(1)

Then the equations in (1) are said to be parametric equations of the curve where ‘t’ is called the

parameter.

Clearly x = at2, y = 2at satisfy the equation y2 = 4ax for all real values of t. Hence the parametric

equations of the parabola y2 = 4ax are x = at2, y = 2at, where t is the parameter.

Also, (at2, 2at) is a point on the parabola y2 = 4ax for all real values of t. This point is also described

as the point ‘t’ on the parabola.

Example -6: The tangents at the points (a 21t , 2at1), (a 2

2t , 2at2) on the parabola y2 = 4ax

are at right angles if (A) t1t2 = -1 (B) t1t2 = 1 (C) t1t2 = 2 (D) t1t2 = -2

Solution: 1t

1.

t

1

21

t1t2 = -1

Hence (A) is the correct answer.

Example -7: If P(a 21t , 2at1) and Q (a 2

2t , 2at2) are two variable points on the curve y2 =

4ax and PQ subtends a right angle at the vertex, then t1t2 is equal to (A) –1 (B) –2 (C) –3 (D) –4

Solution: Slope of OP = 1t

2

Slope of OQ = 2t

2

Given OP OQ 1t

2 .

2t

2 = –1 t1t2 = –4

Hence (D) is the correct answer.

Focal Chord:

Any chord to the parabola y2 = 4ax which passes through the

focus is called a focal chord of the parabola y2 = 4ax.

Let y2 = 4ax be the equation of a parabola and (at2, 2at) a

point P on it. Suppose the coordinates of the other extremity

Q of the focal chord through P are (at12, 2at1).

Then, PS and SQ, where S is the focus (a, 0) have the same

slopes.

aat

0at2

aat

0at22

1

1

2

tt12 – t = t1 t2 - t1 (tt1 + 1)(t1 – t) = 0.

Hence t1 = -1/t, i.e. the point Q is (a/t2, -2a/t).

y2 = 4ax

Q(at12, 2at1)

S(a, 0)

P(at2, 2at)

xA

y

i.e. the extremities of a focal chord of the parabola y2 = 4ax may be taken as the points t and -1/t.

Example -8: Prove that the circle with any focal chord of the parabola y2 = 4ax as its diameter always touches its directrix.

Solution: Let AB be a focal chord. If A is (at2, 2at), then B is

t

a2,

t

a2

.

Equation of the circle with AB as diameter is

(x - at2)

2t

ax + (y - 2at)

t

a2y = 0

For x = -a, this gives

2

222

t

t1a + y2 - 2ay

t

1t - 4a2 = 0 a2

2

t

1t

+ y2 - 2ay(t - 1/t) = 0

[y - a(t - 1/t)]2 = 0, which has equal roots.

x + a = 0 is a tangent to the circle with diameter AB.

Example -9: In the parabola y2 = 4ax, the length of the chord passing through the

vertex and inclined to the axis at an angle /4 is

(A) 4a 2 (B) 2

a4

(C) 2a 2 (D) none of these

Solution: 4

tanat

at22

t = 2

P = (4a, 4a)

OP = 4 2 a.

Hence (A) is the correct answer. X

N

P(at2, 2at)

O

Y

/4

Focal Distance of any Point: The focal distance of any point P (x, y) on the

parabola y2 = 4ax is the distance between the point

P and the focus S, i.e. PS.

Thus the focal distance

= PS = PM =ZN = ZA + AN = a + x Z

M

SX

N

P

A

Y

Position of a point relative to the Parabola:

Consider the parabola : y2 = 4ax. If (x1 , y1) is a given point

and y21 -4ax1 = 0, then the point lies on the parabola. But

when y12 - 4ax1 0, we draw the ordinate PM meeting the

curve in L. Then P will lie outside the parabola if

PM > LM, i.e., PM2 – LM2 > 0

Now, PM2 = y12 and LM2 = 4ax1 by virtue of the coordinates

of L satisfying the equation of the parabola. Substituting

these values in equation of parabola, the condition for P to

lie outside the parabola becomes y12 - 4ax1 > 0.

P(x1, y1)

M

L

X A

y

Interior Exterior

Similarly, the condition for P to lie inside the parabola is y12 -4ax1 < 0.

Example -10:The coordinates of a point on the parabola y2 = 8x, whose focal distance is 4, are

(A)

2,

2

1 (B) (1, 2 2 )

(C) (2, 4) (D) none of these

Solution: Focal distance of a point P (x, y) on y2 = 4ax is (x + a).

4 = x + 2 x = 2

y2 = 8 2 = 16 y = 4

Hence (C) is the correct answer.

1. GENERAL EQUATION OF A PARABOLA Let (h, k) be the focus S and lx + my + n = 0 the equation of the directrix ZM of a parabola. Let (x,y) be the coordinates of any point P on the parabola. Then the relation, PS = distance of P from ZM, gives (x – h)2 + (y – k)2 = (lx + my + n)2 / (l2 + m2)

2(mx ly) 2gx 2fy d 0 This is the general equation of a parabola. Note: From the general equation of the parabola it is clear

that the second-degree terms in the equation of a parabola form a perfect square. The converse is also true, i.e. if in an equation of the second degree, the second-degree terms form a perfect square then the equation represents a parabola, unless it represents two parallel straight lines.

Special case:

Let the vertex be (, ) and the axis be parallel to the x-axis. Then

the equation of parabola is given by (y - )2 = 4a (x – ) which is equivalent to x = Ay2 + By + C

Similarly, when the axis is parallel to the y-axis and vertex be(, ),

the equation of parabola is given by (x - )2 = 4a (y – ) which is equivalent to y = A’x2 + B’x + C’ Example 11: Find the vertex, axis, directrix, tangent at the vertex and the length of

the latus rectum of the parabola 2y2 + 3y - 4x - 3 = 0.

O

Z

M

P(x, y)

S(h, k)

X

Y

X

Y

( , )

X

Y

( , )

Solution: The given equation can be re-written as

32

33x2

4

3y

2

which is of the form Y2 = 4aX.

Hence the vertex is

4

3,

32

33

The axis is 04

3y y = -3/4

The directrix is X + a = 0

x + 32

33 +

2

1 = 0 x =

32

49

The tangent at the vertex is 032

33x

x = -

32

33

Length of the latus rectum = 4a = 2.

INTERSECTION OF A STRAIGHT LINE WITH THE PARABOLA

Points of Intersection of a Straight Line with the Parabola:

Points of intersection of y2 = 4ax and y = mx + c are given by (mx + c)2 = 4ax

i.e. m2x2 + 2x(mc - 2a) + c2 = 0 ……(1)

Since (1) is a quadratic equation, the straight line meets the parabola in two points (real, coincident,

or imaginary). The roots of (1) are real or imaginary according as

{2(mc – 2a)}2 4m2c2 is positive or negative, i.e. according as –amc +a2 is positive or negative, i.e.

according as mc is less than or greater than a.

Note:

When m is very small, one of the roots of equation (1) is very large; when m is equal to zero,

this root is infinitely large. Hence every straight line parallel to the axis of the parabola meets

the curve in one point at a finite distance and in another point at an infinite distance from the

vertex. It means that a line parallel to the axis of the parabola meets the parabola only in one

point.

Length of the Chord:

As in the preceding article, the abscissae of the points common to the straight line y = mx + c and

the parabola y2 = 4ax are given by the equation m2 x2 + ( 2mc – 4a) x + c2 = 0.

If (x1, y1) and (x2, y2) are the points of intersection, then

(x1 - x2)2 = (x1 + x2)2 - 4x1 x2

=

4m

mca16a

2m

24c

4m

22amc4

and (y1 - y2) = m(x1 - x2)

Hence, the required length

= mcaam1m

4 2

2 21

22

21

2

21 xxm1xxyy

TANGENT TO A PARABOLA

Tangent at the Point (x1, y1):

Let the equation of the parabola be y2 = 4ax.

Hence, value of dx

dy at P(x1, y1) is

2a

y1

and the equation of the tangent at P is

y - y1 = 1y

2a(x - x1) i.e. yy1 = 2a(x - x1) + y1

2 yy1 = 2a(x + x1)

Tangent in Terms of m:

Suppose that the equation of a tangent to the parabola y2 = 4ax ……(1)

is y = mx + c ……(2)

The abscissae of the points of intersection of (1) and (2) 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

(mc - 2a)2 = m2c2 ……(3)

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 = 2m

a and y2 = 4ax y =

m

2a.

Thus the point of contact of the tangent y = mx + a/m is

m

2a ,

m

a2

.

Tangent at the Point ‘t’:

Let the equation of the parabola is y2 = 4ax.

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

If the point (x1, y1) (at2, 2at)

Equation of tangent becomes y.2at = 2a(x + at2) yt = x + at2.

Note:

The point of intersection of the tangents at ‘t1’ and ‘t2’ to the parabola y2 = 4ax is

(at1t2, a(t1 + t2)).

Example -12: Two tangents are drawn from the point (-2, -1) to the parabola y2 =

4x. If is the angle between these tangents, then tan equals (A) 3 (B) 1/3 (C) 2 (D) 1/2

Solution: Here a = 1. Any tangent is y = mx + m

1.

It passes through (-2, -1)

2m2 –m –1 = 0

m = 1, 2

1 tan = 3

2

11

2

11

Hence (A) is the correct answer.

Example -13: If the line 2x + 3y = 1 touches the parabola y2 = 4ax, find the length of the latus rectum.

Solution: Equation of any tangent to y2 = 4ax is

m

amxy m2x – my + a = 0

Comparing it with the given tangent 2x + 3y – 1 = 0, we find

1

a

3

m

2

m2

9

2

3

ma,

3

2m

Hence the length of the latus rectum = 4a = 9

8

ignoring the negative sign for length.

Example -14: The angle between tangents to the parabola y2 = 4ax at the point where

it intersects with the line x – y –a = 0 is

(A) 3

(B)

4

(C) 6

(D)

2

Solution: The given chord id a focal chord as it passes through focus (a, 0) and we know that tangents at the extremities of a focal chord meet at right angles on the directrix.

Hence (D) is the correct answer.

Example -15: On the parabola y2 = 4ax, three points E, F, G are taken so that their ordinates are in G.P. Prove that the tangents at E and G intersect on the abscissa of F.

Solution: Let the points E, F, G be (at12, 2at1), (at22, 2at2), (at32, 2at3) respectively.

Since the ordinates of these points are in G.P., t22 = t1t3.

Tangents at E and G are t1y = x + at12 and t3y = x + at32.

Eliminating y from these equation, we get

x = at1t3 = at22, which is the abscissa of F.

Equation of the Tangents from an External Point:

Let y2 = 4ax be the equation of a parabola and (x1, y1) an external point P. Then, equations of the

tangents from (x1, y1) to the given parabola are given by

SS1 = T2, where S = y2 - 4ax, S1 = y12 - 4ax1, T = yy1 - 2a(x + x1)

Chord of Contact:

Equation to the chord of contact of the tangents drawn from a point (x1, y1), to the parabola

y2= 4ax is T= 0, i.e. yy1 - 2a(x+x1) =0.

Equation of a Chord with Midpoint (x1, y1):

The equation of the chord of the parabola y2= 4ax with mid point (x1, y1) is T= S1

i.e. yy1-2a(x+x1)= y12-4ax1 Or yy1 - 2ax = y1

2 - 2ax1 .

Example -15: Find the equation of the chord of the parabola y2 = 12x which is bisected at the point (5,

–7).

Solution: Here (x1, y1) = (5 –7) and y2 = 12x, a = 3

The equation of the chord is S1 = T

or 2

1y - 4ax1 = yy1 – 2a (x + x1)

or (–7)2 – 12.5 = y(–7) – 6(x+5)

6x + 7y + 19 = 0.

NORMAL TO THE PARABOLA

Normal at the Point (x1, y1 ):

The equation of the tangent at the point (x1, y1) is yy1 = 2a(x + x1). Since the slope of

tangent = 2a/y1 , slope of normal is -y1/ 2a . Also it passes through (x1, y1).

Hence its equation is y - y1 = 11 xx

2a

y . . . . . (i)

Normal in Terms of m:

In equation (i), put m2a

y1 so that y1 = -2am and x1 = 22

1 am4a

y , then the equation becomes y =

mx - 2am - am3 . . . . . (ii)

where m is a parameter. Equation (ii) is the normal at the point (am2, -2am) of the parabola.

Notes:

If this normal passes through a point (h, k), then k = mh – 2am - am3.

For a given parabola and a given point (h, k) , this cubic in m has three roots say m1,

m2, m3 i.e. from (h, k) three normals can be drawn to the parabola whose slopes

are m1, m2, m3 . For the cubic, we have

m1+ m2 + m3 = 0

m1 m2 +m2 m3 +m3 m1 = (2a-h) /a

m1 m2 m3 = - k/a

If we have an extra condition about the normals drawn from a point (h, k) to a given parabola

y2 =4ax then by eliminating m1, m2, m3 from these four relations between m1, m2, m3, we

can get the locus of (h, k).

Since the sum of the roots is equal to zero, the sum of the ordinates of the feet of the

normals from a given point is zero.

Normal at the Point ‘t’:

Equation of the normal to y2 = 4ax at the point (x1, y1) is y – y1 = –a2

y1 (x – x1)

If (x1, y1) (at2, 2at)

Equation of normal becomes

y – 2at = –a2

at2(x – at2) y = –tx + 2at + at3.

Notes: If normal at the point t1 meets the parabola again at the point t2, then t2 = -t1 –2/t1.

Point of intersection of the normals to the parabola y2 = 4 ax at (at12, 2at1) and

(at22, 2at2) is (2a + a(t12 + t2

2 + t1t2),– at1t2(t1+ t2)) .

Example -16: Find the locus of the foot of the perpendicular drawn from the vertex on a tangent to the parabola y2 = 4ax.

Solution: Any tangent of slope m to the parabola y2 = 4ax is

y = mx + a/m . . . (1)

The perpendicular from the vertex on (1) is

x + my = 0 . . . . (2)

By eliminating m between (1) and (2) we obtain the required locus as xy2

+ x3 + ay2 = 0

Example -17: If the normal at the point (at12, 2at1) meets the parabola y2 = 4ax again at the point (at2

2,

2at2), prove that t2 = –1

1t

2t

Solution: The equation of the normal at (at12, 2at1) is

y = –t1x + 2at1 + at13.

Since the normal passes through the point (at22, 2at2),

we have, 2at2 = -t1. at22 +2at1 + at13

2a (t1 – t2) = at1 t22 – at13

= at1 )tt( 21

22 = at1(t2 – t1) (t2 + t1)

2 = –t1 (t2 + t1), as t1 t2

t2 + t1 = 1t

2 t2 = –t1

1t

2

Example -18: The normal at any point P to a parabola y2 = 4ax meets its axis at G. Q is another point on the parabola such that QG is perpendicular to the axis of the parabola. Prove that QG2 – PG2 = constant.

Solution: Let P be the point (at2, 2at)

Equation of normal at P is

y = –tx + 2at + at3 ……(1)

Equation (1) meets the x-axis at G whose

coordinates are (2a + at2, 0)

PG2 = 4a2 + 4a2t2 ……(2)

Since QG is perpendicular to the axis, the

abscissa of Q is the same as that of G and its

ordinate is QG.

Point Q is (2a + at2, QG)

But Q lies on the parabola y2 = 4ax.

y2 = 4ax

G

P(at2, 2at)

(0, 0)

Q

QG2 = 4a(2a + at2) = 8a2 + 4a2t2 ……(3)

QG2 – PG2 = 4a2 = constant by (2) and (3).

Subtangent and Subnormal:

Let the tangent and normal at any point

P (x1, y1) on the parabola y2 = 4ax meet the axis

in T and G respectively. Then PT is called the

length of the tangent at P and PG is called the

length of the normal at P.

NT is called the subtangent and NG the

subnormal at P. The coordinates of T and G can

be easily found by putting x = 0 in the equations

of the tangent and normal at P. It is evident from

figure that

Subtangent = NT = twice the abscissa of P

Subnormal = NG = 2a = semilatus rectum

y2 = 4ax

G

P(x1, y1)

A

Q

X T (–x1, 0) S

(a, 0) (x1, 0) N (x1 + 2a, 0)

X

yy1 = 2a(x + x1)

Y

Y

Example -19: If P is a point on the parabola y2 = 4ax such that the subtangent and subnormal at P are equal, find the coordinates of P.

Solution: Let P (x, y) be the required point.

Length of subtangent = twice the abscisse of P = 2x

Length of subnormal = 2a

Since subtangent = subnormal, 2x = 2a x = a y2 = 4a2 y = 2a

The required points are (a, 2a) and (a, –2a).

PROPERTIES OF THE PARABOLA

P

y2 = 4ax

S (Focus) xA

T

M

y

K

Z

MP

N G

(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.

In the given figure MPT = TPS :

Similarly the normal at any point on a parabola bisects the angle between the focal chord

and the line parallel to the axis through that point.

(ii) The portion of a tangent to a parabola cut off between the directrix and the

curve subtends a right angle at the focus.

In the given figure SP is perpendicular to SK i.e. KSP = 900 .

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

directrix.

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

the tangent at the vertex.

Pole and Polar

To find the equation of the polar of the point (x1, y1) with respect to the parabola y2 = 4ax.

Let Q and R be the points in which any chord drawn through the point P, whose coordinates are (x1,

y1), meets the parabola.

Let the tangents at Q and R meet in the point whose coordiantes are (h, k).

R’T’

P

Q

RT

(h, k)

(x1, y1)

R’

T’

P

Q’

R

T(h, k)

(x1, y1)

Q

We require the locus of (h, k).

Since QR is the chord of contact of tangents from (h, k) its equation is ky = 2a (x + h)

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

ky1 = 2a (x 1+ h) ….(i)

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

yy1 = 2a(x + x1) ….(ii)

Hence (ii) is the equation to the polar of (x1, y1).

Example -20:Prove that the locus of poles of focal chord of the parabola y2 = 4ax is

the directrix.

Solution: Let (h, k) be the pole. Then the equation of the chord is ky = 2a(x + h).

Since it is a focal chord, it passes through the focus (a, 0).

2a(a + h) = 0 or a + h = 0

Hence locus of pole (h, k) is

x + a = 0, which is the directrix.