Mathematics

Session

Parabola Session 3

Session Objective

1. Number of Normals Drawn From a Point2. Number of Tangents Drawn From a Point3. Director circle4. Equation of the Pair of Tangents5. Equation of Chord of Contact6. Equation of the Chord with middle point at

(h, k)7. Diameter of the Parabola8. Parabola y = ax2 + bx + c

Number of Normals Drawn From a Point (h,k)

Parabola be y2 = 4ax let the slope of the normal be m, then its equation is given by y = mx – 2am – am3 if it passes through (h,k) then

k = mh – 2am – am3

i.e. am3 + (2a – h)m + k = 0

This shows from (h,k) there are three normals possible (real/imaginary) as we get cubic in m

Observations from am3 + (2a – h)m + k = 0

1. At least one of the normal is real as cubic equation have atleast one real root

2. The three feet of normals are called Co-Normal points given by (am1

2, –2am1 ), (am22, –2am2) and

(am32, –2am3) where mi’s are the roots of the

given cubic eqn

3. Sum of the ordinates of the co-normal points = –2a (m1 + m2 + m3) = 0

4. Sum of slopes of normals at co-normal points = 0

5. Centroid of triangle formed by co-normal points lies on axis of the parabola.

Observations from am3 + (2a – h)m + k = 0

1 2 3

1 2 2 3 3 1

1 2 3

6. m +m +m = 0

2a- hm m +m m +m m =

a-k

m m m =a

7. Thus we have following different cases arises:

• 3 real and distinct roots m1, m2, m3 or m1, m2, –m1–m2

• 3 real in which 2 are equal m1, m2, m2 or –2m2, m2, m2

• 3 real, all equal m1, m1, m1 or 0, 0, 0 k = 0 , h = 2a

• 1 real, 2 imaginary m1, i ( 0)

Number of Tangents Drawn From a Point (h,k)

Parabola be y2 = 4ax let the slope of the tangent be m, then its equation is given by y = mx + a/m if it passes through (h,k) then

k = mh + a/m

i.e. hm2 – km + a = 0

This shows from (h,k) there are two tangents possible (real/imaginary) as we get quadratic in m

Observations from hm2 – km + a = 0

Discriminant = k2 – 4ah = S1

1. S1 > 0 Point is outside parabola: 2 real & distinct tangents

2. S1 = 0 Point is on the parabola: Coincident tangents

3. S1 < 0 Point is inside parabola: No real tangent

4. m1 + m2 = k/h , m1m2 = a/h

Director Circle

Locus of the point of intersection of the perpendicular tangents is called Director Circle

hm2 – km + a = 0

m1m2 = a/h = –1

h = –a i.e. locus is x = –a

Hence in case of parabola perpendicular tangents intersect at its directrix.

Director circle of a parabola is its directrix.

Equation of the Pair of Tangents

Parabola be y2 = 4ax then equation of pair of tangents drawn from (h,k) is given by

SS1 = T2

where S y2 – 4ax, S1 k2 – 4ah and T ky – 2a(x + h)

Pair of Tangents:

(y2 – 4ax)(k2 – 4ah) = (ky – 2a(x + h))2

O

(h, k)x

y

Equation of Chord of Contact

Parabola be y2 = 4ax then equation of chord of contact of tangents drawn from (h,k) is given by

T = 0

where T ky – 2a(x + h)

Chord of Contact is:

ky = 2a(x + h)

O

(h, k)x

y

Equation of the Chord with middle point at (h, k)

Parabola be y2 = 4ax then equation of chord whose middle point is at (h,k) is given by

T = S1

where T ky – 2a(x + h) and S1 k2 – 4ah

Chord with middle point at (h,k) is:

ky – 2a(x + h) = k2 – 4ah

i.e. ky – 2ax = k2 – 2ah

O(h, k)

x

y

Diameter of the Parabola

Diameter: Locus of mid point of a system of parallel chords of a conic is known as diameter

Ox

y

y = —2am

y = 4 ax2

y = mx + c

Let (h, k) be the mid point of a chord of slop e m then its equation is given by

ky – 2ax = k2 – 2ah

if its slope is m then

2a= m

kLocus is

2ay =

m

Class Exercise

Class Exercise - 1

Find the locus of the points from whichtwo of the three normals coincides.

Solution

Let (h, k) be the point and m be the slope

of the normal then 3am 2a h m k 0

As two normal coincides

Let m1 =m2 then 1 3 32m m 0 m 2m

and 21 3

km m

a

3 31 1

k k2m m

a 2a

Solution contd..

31 1As am 2a h m k 0

3 33 3

1 1am k 2a h m

3

33k k2a h

2 2a

3227ak 4 h 2a

32 locus of (h, k) is 27ay 4 x 2a

Class Exercise - 2

Three normals with slopes m1, m2

and m3 are drawn from a point Pnot on the axis of the parabola y2 = 4x.If results in the locus of Pbeing a part of the parabola, find thevalue of

1 2m m ,

.

Solution

Let P be (h, k) then mi’s are given by

3m 2 h m k 0 ...(i) (as a 1)

1 2 3m m m k 1 2 3k

m m m

m3 satisfies (i) 3

3

2 hkk 0

k 0 as (h, k) doesn't lies on axis

2 2 3k 2 h 0 2 2 3 2k h 2

Solution contd..

locus of (h, k) is 2 2 3 2y x 2

If it is a part of y2 = 4x

2 3 24 and 2 0

22 and 2 0 0, 2

hence 2

Class Exercise - 3

Find the locus of the middle pointsof the normal chords of the parabolay2 = 4ax.

Solution

Let (h, k) be the middle point then

its equation is given by T = S1 i.e.

ky – 2a (x + h) = k2 – 4ah

2a 2ahy x k .

k k

If it is also the normal of y2 = 4ax then compare it with 3y mx am 2am

Solution contd..

32ah 2a 2a

k a 2ak k k

32ah 2a 2a

k a 2ak k k

32ah 2a 2a

k a 2ak k k

2 2 2 4 42ahk 4a k 8a k 2 4 42ak h 2a 8a k

2 4 4locus of (h, k) is 2ay x 2a 8a y

Class Exercise - 4

Find the locus of the point ofintersection of the tangents at theextremities of chord of y2 = 4ax whichsubtends right angle at its vertex.

Solution

Let (h, k) be the point of intersection of

tangents then its chord of contact is given

by T = 0 i.e.

ky – 2a (x + h) = 0 ...(i)

Pair of lines are given by homogenising y2 = 4ax

using (i) 2 ky 2axy 4ax

2ah

Now according to the question pair of lines

joining origin to the point of intersection of

(i) with the parabola are at right angles.

Solution contd..

2hy 2x ky 2ax

4ax2 – 2kxy + hy2 = 0

Pair of lines are perpendicular if

4a + h = 0

hence locus of (h, k) is x + 4a = 0

Alternative:

Let be the extremities

of the chord. As chord subtends right angle at the

vertex we have

2 21 1 2 2at , 2at and at , 2at

1 21 22 2

1 2

2at 2at. 1 t t 4

at at

Solution contd..

Point of intersection of tangents

at these point is 1 2 1 2at t , a t t

1 2now t t 4

point becomes (–4a, a (t1 + t2)) and its locus is x + 4a = 0

Class Exercise - 5

Find the equation of the diameter of theparabola given by 3y2 = 7x, whose systemof parallel chords are y = 2x + c.

Solution

Let (h, k) be the middle point of the

chord then its equation is given by T = S1

273ky x h 3k 7h

2

Its slope is 2 7

22.3k

7k

12

7i.e. locus of (h, k) is y

12

Solution contd..

Alternative:

2ay is the equation of diameter

m

72. 73.4y

2 12

Class Exercise - 6

Three normals to the parabola y2 = xare drawn through a point (c, 0), then

(a) (b)

(c) (d) None of these

1c

4

1c

2

1c

2

Solution

3am 2a h m k 0

31 1m 2. c m 0

4 4

2m m 2 4c 0

above equation have 3 real roots if

2 – 4c < 0

i.e. 1

c2

Hence,answer is (c).

Class Exercise - 7

The mid point of segmentintercepted by the parabola x2 =6yfrom the line x – y = 1 is ___.

Solution

Let the mid point be (h, k), therefore, its

equation is given by

hx – 3 (y + k) = h2 – 6k

or hx – 3y = h2 – 3k

2h 3 h 3k1 1 1

h = 3, k = 2

Hence (3, 2) is the mid point of the segment.

Class Exercise - 8

Draw y= –2x2 + 3x + 1.

Solution

2 3y 2 x x 1

2

23 9

2 x 14 16

23 9

2 x 14 8

23 17

2 x4 8

23 1 17

x y4 2 8

x = 34—

34— 17

8—, 1

y

o x

Class Exercise - 9

Find the locus of the point of intersectionof tangents to y2 = 4ax which includes anangle between them.

Solution

Let (h, k) be the point of intersection

then equation of pair of tangents are

given by SS1 = T2

22 2i.e. y 4ax k 4ah ky 2a x h

2 2 2k 4ah y 4a k 4ah x

2 2 2 2 2 2 2k y 4a x 4a h 8a hx 4akxy 4ahky

2 2ax kxy hy ... 0

Solution contd..

2k

2 ah2

tana h

2k 4aha h

locus of (h, k) is

22 2y 4ax tan x a

Class Exercise - 10

Find the coordinates of feet of thenormals drawn from (14, 7) to theparabola y2= 16x + 8y.

Solution

Let the foot of the normal be ,

then 2

2 816 8

16

tangent at is given by ,

y 8 x 4 y

8Slope of tangent

4

4Slope of normal becomes

8

Solution contd..

equation of normal at , is

4 x 8y 4 8

which also passes through (14, 7)

4 14 56 4 8

26 84 16

0 or 8 4 96

2i.e. 12 64 0 16 4 0

0, 16, 4 Corresponding 0, 8, 3

Feet of normals are (0, 0), (8, 16) and (3, –4)