Transcript Document

Mathematics
Session
Parabola Session 3
Session Objective
1.
2.
3.
4.
5.
6.
Number of Normals Drawn From a Point
Number of Tangents Drawn From a Point
Director circle
Equation of the Pair of Tangents
Equation of Chord of Contact
Equation of the Chord with middle point at
(h, k)
7. Diameter of the Parabola
8. 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 (am12, –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
6. m1 + m2 + m3 = 0
m1m2 + m2m3 + m3m1 =
2a - h
a
-k
a
7. Thus we have following different cases arises:
m1m2m3 =
•
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:
4. m1 + m2 = k/h , m1m2 = a/h
No real tangent
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
y
SS1 = T2
where S  y2 – 4ax, S1  k2 – 4ah and
T  ky – 2a(x + h)
(h, k)
O
Pair of Tangents:
(y2 – 4ax)(k2 – 4ah) = (ky – 2a(x + h))2
x
Equation of Chord of Contact
Parabola be y2 = 4ax then equation of chord
of contact of tangents drawn from (h,k) is
given by
y
T=0
where T  ky – 2a(x + h)
Chord of Contact is:
ky = 2a(x + h)
(h, k)
O
x
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
y
Chord with middle point at (h,k) is:
ky – 2a(x + h) = k2 – 4ah
i.e. ky – 2ax = k2 – 2ah
(h, k)
O
x
Diameter of the Parabola
Diameter: Locus of mid
point of a system of parallel
chords of a conic is known
as diameter
y
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
k
y2 = 4 ax
Locus is
y =
O
2a
y=—
m
y = mx + c
2a
m
x
Class Exercise
Class Exercise - 1
Find the locus of the points from which
two of the three normals coincides.
Solution
Let (h, k) be the point and m be the slope
3
of the normal then am   2a  h m  k  0
As two normal coincides
Let m1 =m2 then 2m1  m3  0
2
and m1 m3 

 2m13
k
a
k

a

m13
k

2a
 m3  2m
Solution contd..
As am13   2a  h m1  k  0

am13
k

3
   2a  h m13
3
3
3 k
 3k 
      2a  h
2a
 2
27ak 2  4 h  2a 
3
 locus of (h, k) is 27ay  4  x  2a 
2
3
Class Exercise - 2
Three normals with slopes m1, m2
and m3 are drawn from a point P
not on the axis of the parabola y2 = 4x.
If m1 m2  , results in the locus of P
being a part of the parabola, find the
value of .
Solution
Let P be (h, k) then mi’s are given by
m3   2  h m  k  0 ...(i) (as a  1)
m1 m2 m3  k
m3 satisfies (i)
k
m1m2    m3 


k3
3
2  h


k 0

k  0 as (h, k) doesn't lies on axis
 k 2   2  h   2  3  0
 k 2  2h  3  22
Solution contd..
locus of (h, k) is y   x    2
2
2
3
2
If it is a part of y2 = 4x
 2  4 and 3  22  0
  2 and 2    2  0    0, 2
hence   2
Class Exercise - 3
Find the locus of the middle points
of the normal chords of the parabola
y2 = 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
2ah
y
xk 
.
k
k
If it is also the normal of y2 = 4ax then compare it with
y  mx  am3  2am
Solution contd..
3
2ah
 2a 
 2a 
 k
 a    2a  
k
 k 
 k 
3
2ah
 2a 
 2a 
 k
 a    2a  
k
 k 
 k 
3
 k
2ah
 2a 
 2a 
 a    2a  
k
 k 
 k 
2ahk 2  4a2k 2  8a4  k 4
2ak 2 h  2a   8a4  k 4
 locus of (h, k) is 2ay2  x  2a   8a4  y 4
Class Exercise - 4
Find the locus of the point of
intersection of the tangents at the
extremities of chord of y2 = 4ax which
subtends 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)
Now according to the question pair of lines
joining origin to the point of intersection of
(i) with the parabola are at right angles.
Pair of lines are given by homogenising y2 = 4ax
ky  2ax 
using (i) y 2  4ax 

 2ah 
Solution contd..
hy2  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:



2
2
Let at1 , 2at1 and at2, 2at 2

be the extremities
of the chord. As chord subtends right angle at the
vertex we have
2at1 2at2
. 2  1  t1 t2  4
2
at1 at2
Solution contd..
Point of intersection of tangents

at these point is at1 t 2 , a  t1  t 2 

now t1 t 2  4
 point becomes (–4a, a (t1 + t2)) and its locus is x + 4a = 0
Class Exercise - 5
Find the equation of the diameter of the
parabola given by 3y2 = 7x, whose system
of 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
7
3ky   x  h   3k 2  7h
2
Its slope is 2 
7
2
2.3k
7
 k
12
7
i.e. locus of (h, k) is y 
12
Solution contd..
Alternative:
y
y
2a
is the equation of diameter
m
2.
7
3.4  7
2
12
Class Exercise - 6
Three normals to the parabola y2 = x
are drawn through a point (c, 0), then
(a) c 
1
4
(b) c 
1
2
(c) c 
1
2
(d) None of these
Solution
am3   2a  h m  k  0
1 3  1

m   2.  c  m  0
4
 4



m m2  2  4c  0
above equation have 3 real roots if
2 – 4c < 0
1
i.e. c 
2
Hence,answer is (c).
Class Exercise - 7
The mid point of segment
intercepted by the parabola x2 =6y
from 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
h 3 h2  3k
 

1 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
3 

y  2  x2  x   1
2 

2

3
9
 2   x      1
4
16 


2
3
9

 2  x     1
4
8

y
1
3 17
—, —
4 8
2
3  17

 2  x   
4
8

2
3
1
17 

x


y





4
2
8




x
o
3
—
x=4
Class Exercise - 9
Find the locus of the point of intersection
of tangents to y2 = 4ax which includes an
angle  between them.
Solution
Let (h, k) be the point of intersection
then equation of pair of tangents are
given by SS1 = T2



2
2
2
k

4ah
y

4a
k


  4ah x
i.e. y  4ax k  4ah  ky  2a  x  h
2
2
2
 k 2 y2  4a2 x2  4a2h2  8a2hx  4akxy  4ahky
ax2  kxy  hy2  ...  0
Solution contd..
2
k
2    ah
2
 tan  
ah
k2  4ah

ah
 locus of (h, k) is
y  4ax  tan   x  a
2
2
2
Class Exercise - 10
Find the coordinates of feet of the
normals drawn from (14, 7) to the
parabola y2= 16x + 8y.
Solution
Let the foot of the normal be  ,  
2  8
then   16  8   
16
2
tangent at  ,   is given by
y  8  x     4  y   
8
Slope of tangent 
4
4
 Slope 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
6 2  8


4
16
   0 or
i.e. 2  12  64  0

   0, 16,  4
  8   4     96
  16    4   0
Corresponding   0, 8, 3
 Feet of normals are (0, 0), (8, 16) and (3, –4)