Chap-12 - Conic Sections.docx

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12
P
S
Conic Sections
Sections of cones, equations of conic sections in standard forms. Condition
for y = mx + c to be a tangent and point (s) of tangency
Conic sections, namely a point, pair of straight lines, circle, ellipse, parabola and
hyperbola are called so because they can be obtained when a cone (or double
cone) is cut by a plane.
The mathematicians associated with the study of conics were Euclid, Aristarchus
and Apollonius. Most of the objects around us and in space have shape of conic-
sections. Hence study of these becomes a very important tool for present
knowledge and further exploration.
Conic-sections as Sections of Right Circular Cone(s)
1. When a double right circular cone is cut by a plane parallel to base at the
common vertex, the cutting profile is a point.
C H A P T E R
CHAPTER
INCLUDES :
 Conic sections
 General equation of
conics
 Parabola
 Ellipse
 Hyperbola
 Solved examples
2. When a double right circular cone is cut by plane, parallel to its common axis,
the cut profile is hyperbola
3. When a right circular cone is cut by a plane parallel to its base the cutting
profile is a circle.
Syllabus

JEE main Conic Sections
 
4. When a right circular cone is cut by any plane through its vertex, the cutting profile is a pair of straight lines
through its vertex
5. When a right circular cone is cut by a plane parallel to a generator of cone, the cutting profile is a parabola.
6. When a right circular cone is cut by a plane which is neither parallel to any generator / axis nor parallel to
base, the cutting profile is an ellipse.
Hence a point, a pair of intersecting straight lines, circle, parabola, ellipse and hyperbola, all are conic-
sections. All the conic sections are plane or two dimensional curves.
The conic section is the locus of a point which moves such that the ratio of its distance from a fixed point
(focus) to the distance from a fixed straight line (directrix) is constant (e), e is called eccentricity of conic
i.e.,
PS
 e
PK
e = 1 for parabola
e < 1 for ellipse
e > 1 for hyperbola
A line through focus and perpendicular to directrix is called
- axis. The vertex of conic is that point where the curve
intersects the axis.
PS
 e  PS2  e2PK 2
PK
K
directrix
Ax+By+C=0
 Ax  By  C 
2
or (x  )2  (y  )2  e2  
 A2
 B2 
P(x, y)
S(a, b)
focus
axis

Conic Sections JEE main
2
y
S
(a, 0)
Note : (i) For parabola PN2
= 4AS.AN.
(ii) Two parabolas are said to be equal when their latus recta are equal.
Simplification shall lead to the equation of the form
ax2
+ by2
+ 2hxy + 2gx + 2fy + c = 0
So equation of a conic is the general equation of second degree. If
 = abc + 2fgh – af2
– bg2
– ch2
if   0  a pair of straight lines
if   0 and h = 0 and a = b  a circle
if   0 and h2
= ab  a parabola
if   0 and h2 < ab  an ellipse
if   0 and h2
> ab  a hyperbola
PARABOLA
The parabola shown has the equation y2 = 4ax. z
For this parabola M
(i) Vertex A is (0, 0)
(ii) Focus S is (a, 0)
(iii) Equation of directrix zz´ is x = – a.
(iv) Equation of axis AX is y = 0
K
Focal chord : A chord of a parabola which passes through
the focus is called as focal chord.
Focal distance : The distance, of a point on the parabola, from z
the focus is called as focal distance of the
point.
y
P
L
N x
A S(a, 0)
L
Latus rectum : The double ordinate LSL´ through the focus is the latus rectum. Its equation is x = a and
its length is 4a units.
The equation of the parabola, whose focus is (3, –4) and directrix is the line y + x = 2 is
Solution :
PS2
= PM2
Gives
 x  y  2 
2
(x  3)2  (y  4)2   
 
or 2x2
+ 2y2
+ 18 + 32 – 12x + 16y = x2
+ y2
+ 4 – 4x – 4y + 2xy
or x2
+ y2
– 2xy – 8x + 20y + 46 = 0
Equations of Parabola in Standard Forms
(i) y2
= 4ax
Focus : (a, 0)
Directrix : x = – a
Vertex : (0, 0)
Axis : y = 0
x
x = –a
Illustration 1 :

y
S
y
S
(–a, 0)
x = a
y
S (0, a)

Illustration 2 :
(ii) y2
= – 4ax
Focus : (–a, 0)
Directrix : x = a x
Vertex : (0, 0)
Axis : y = 0
(iii) x2
= 4ay
Focus : (0, a)
Directrix : y = – a
Vertex : (0, 0)
x
Axis : x = 0
(iv) x2 = – 4ay
Focus : (0, – a)
Directrix : y = a x
Vertex : (0, 0)
Axis : x = 0
Find the focus, vertex, length of latus rectum and directrix of parabola x2
+ 4x + 4y + 16 = 0
Solution :
(x + 2)2 = –4y – 12 = –4 (y + 3)
 X2 = –4Y  where X = x + 2 and Y = y + 3

or x = (X – 2) and y = (Y – 3)
a = 1, hence latus rectum = 4a = 4 units in length.
In new coordinates, focus : (0, –1), vertex (0, 0), directrix  Y = 1
In original co-ordinates, using x = X – 2 and y = Y – 3
focus (–2, –4), vertex (–2, –3)
and directrix  y + 3 = 1
or, y + 2 = 0
Parametric co-ordinates
The parametric co-ordinates of any point on the parabola y2
= 4ax are given by :
x = at2
, y = 2at; t is the parameter
Position of a point with respect to the parabola y2
= 4ax.
A point P(x1, y1) lies inside, on or outside the parabola y2 = 4ax according as
y2  4ax  0,= 0 or > 0
1 1


1 1 2 2 1 2
1 1 1 1

Illustration 3 :
Equations of a Chord
Chord joining (x1, y1) and (x2, y2) on parabola y2 = 4ax is
y(y1 + y2) = 4ax + y1y2
Chord joining (at 2
, 2at ) and (at 2
, 2at ) is  This chord will be a focal chord is t t = –1
y(t + t ) = 2(x + at t ) 
1 2 1 2 
Length of chord y = mx + c to the parabola y2
= 4ax
Length =
Length of focal chord joining (at 2
, 2at ) and (at 2
, 2at ) = (t – t )2
1 1 2 2 2 1
 12
Length of focal chord through (at2
, 2at) on y2
= 4ax = at 
t

 

Chords and Tangents
Corresponding to any point P(x1, y1) we define the following expressions for the conic
viz, S  ax2  2hxy  by 2  2gx  2fy  c  0
T  axx1  h(xy1  x1y)  byy1  g(x  x1)  f (y  y1)  c
S1  ax2  2hx y  by2  2gx  2fy  c
(i) Equation of tangent at P(x1, y1) to y2
= 4ax is T  0
i.e. yy1 = 2a(x + x1)
(ii) Equation of tangent in slope form is y  mx 
a
m
 a
,
2a 
Its point of contact is 
m2
m


(iii) The line y = mx + c is a tangent to y2
= 4ax if c 
a
m
(iv) If the line y = mx + c intersects the parabola y2 = 4ax then, length of the chord intercepted is
x2  x1
where x1 and x2 are abscissa of points of intersection and arise as roots of the equation
m2x 2  2x(mc  2a)  c2  0
Find the equation of common tangents to the parabolas y2
= 4ax and x2
= 4by.
Solution :
The general equation of tangent to y2 = 4ax is y  mx 
a
m
…(1)
2  a 
 Let (1) be a tangent to x2
= 4by ; then, x  4bmx  
m must have coincident roots
16ab
 

 a 1 3
 16b2m2   0 ; m    

m  b 

1 1 2 2
 (1) becomes, b3  y  a3  x  a3  b3  0
4
m2
1 m2 a(a  mc )
1 m2
1 1

1 2
1

1
1
1
1 2
1 1 1

Illustration 4 :
(i) Equation of pair of tangents
Through any given point P(x1, y1) there pass, in general, two tangents to y2
= 4ax. When P is external to
the parabola, the joint equation to the pair of tangents from P is SS1  T 2
(ii) Points of intersection of two tangents
The tangents to the parabola S  y2  4ax  0 drawn at the points (at2, 2at1) and (at 2, 2at2 ) intersect
at the point at1t2,
(iii) Chord of contact
a(t1  t2 ).
Equation of chord of contact of the point P(x1, y1) (i.e. of tangents from P) with respect to the parabola
y2 = 4ax is T  0
(iv) Chord with middle point P(x1, y1) has equation : T  S1
Find the locus of the mid-point of the chord of the parabola y2
= 4ax, which subtends right angle at the vertex.
Solution :
Let the mid point of the chord be (x1, y1).
Equation of the chord is T = s1.
i.e. yy1 = 2a(x + x1) = y 2 – 4ax1..................................................(i)
Equation of the parabola is y2
= 4ax........................... (ii)
Maping (ii) homogeneous with the help of (i),
2  yy1 – 2a(x  x1) 
y  4ax


y 2 – 4ax1

 y2
(y 2
– 4ax ) = 4ay xy – 8a2
x2
– 8a2
xx
1 1 1 1
 8a2x2 – 4ay1xy + (y 2 – 4ax )y2 + 8a2x x = 0
Q Chords subtends right angle at the vertex (0, 0)
 a + b = 0  8a2
+ y 2
– 4ax1 = 0
 y 2
= 4ax1 – 8a2
Required locus  y2
= 4a(x – 2a)
NORMALS TO THE PARABOLA
(i) Equation of normal to the parabola y2
= 4ax at the point P(x , y ) is y  y  
y1
(x  x )
1 1 1
2a 1
(ii) Equation of normal to the parabola y2
= 4ax in slope form is y  mx  2am  am3
Its foot is given by (am2
, –2am)
(iii) Equation of normal to the parabola y2
= 4ax at (at2
, 2at) is y  tx  2at  at3
(iv) The normals at the points (at2, 2at1) and (at2, 2at2 ) intersect at the point
(2a  a(t2  t t  t2 ), –at t (t + t ))
1 1 2 2 1 2 1 2
(v) Through a given point, in general, three normals can be drawn to a parabola
(vi) The sum of the slopes of the normals drawn from a given point to a parabola, is zero.
(vii) The sum of the ordinates of the feet of the normals drawn from a given point to a parabola is zero.
(viii) If the normal at the point P(at2, 2at ) meets the parabola at Q(at2, 2at ), then t  t 
2
1 1 2 2 2 1
1
t

Illustration 5 :
K y
K
L B
1 L1
M
P
M 
Z A S C S N A Z
L2 B L2
If three normals to the parabola y2
= x are drawn through a point (c, 0), then show that c 
1
.
2
Solution :
Any normal to the parabola y2
= 4ax, is y = mx – 2am – am3
.
 In the parabola, y2
= x, a 
1
.
4
 Equation of the normal to the parabola y2 = x is
y 
1 1 3
mx – m – m
2 4
. This passes through (c, 0)
0 
1 1 3
 cm – m – m
2 4
 m(4c – 2 – m2
) = 0
 m = 0 or m2
= 4c – 2
 n = 0 or m  2 c –
1
2
1
 For the real values of m, c  .
2
ELLIPSE
An ellipse is the locus of a point which moves such that its distance from a fixed point is e (< 1) times its
distance from a fixed straight line. Unlike a parabola an ellipse has
(i) two foci
(ii) two directrices – one corresponding to each focus.
(iii) a center – the point such that all chords through it are bisected at it.
(iv) two axes – major and minor axes
(v) two vertices
(vi) two latus recta
Thus, the ellipse shown, has the equation
x2

y 2



a2
b2
(i) Center is C(0, 0)
(ii) Vertices are A(–a, 0) and A´(a, 0)
(iii) Foci are S´(ae, 0) and S(–ae, 0)
x
(iv) Directrix K´M´Z´ corresponds to focus S´
and has equation x 
a
.
e
Directrix KMZ corresponds to focus S
and has equation x  
a
e
 
2b2
(v) L1L2 and L1 L2 are double ordinates through the foci and are called latus rectum. Each has length
a
.
(vi) AA´ is the major axis and has length 2a. BB´ is the minor axis and has length 2b where a > b and
b2  a2 (1 e2 )
Note : 1. An ellipse is the locus of a point which moves such that the sum of its distances from two
fixed points remains constant. In fact, PS + PS´ = 2a for all points P(x, y) on the ellipse.
2. The constant in the above definition should be greater than the distance between the fixed
points. This definition is called the physical definition of the ellipse.
1

The equation of an ellipse, referred to its axes as the axes of coordinates, with foci (± 2, 0) and latus rectum
is 6 units.
Solution :
ae = 2
centre (0, 0);
2b2


a
6  b2  3a
2 b2 2 2 2 2
e  1  a e
a2
 a  b s
 4  a2  3a
 (a  4)(a 1)  0
a = 4, – 1
 a2  3a  4  0
a cannot be negative, hence a = 4
 b  2 3
So the equation of ellipse is
x2

y 2

1
16 12
Auxillary Circle
The circle which is described on the major
axis AA´ of an ellipse as diameter is called
auxillary circle of the ellipse.
Equation of ellipse in parametric form
x = a cos  , y = b sin  ; 0   < 2
y
Q
B P
 x
A N A
B
where  is the parameter (as shown in the figure above) and is called as eccentric angle corresponding to the
point P on the ellipse. The point Q has co-ordinates (a cos , a sin ). Further, PN

b
.
x2

y 2


 
QN a
Position of a point with respect to the ellipse 1
a2
b2
(x)2

(y )2


 
The point R(x´, y´) lies within, upon or outside the ellipse according as 2
positive.
a
1 is negative, zero or
b2
The curve represented parametrically by x = t2 + t + 1 and y = t2 – t + 1 is
(1) A pair of straight lines (2) An ellipse (3) A circle (4) A parabola
Solution :
x + y = 2t2
+ 2
x – y = 2t
 x  y 2
 x  y  2

  2
2 
2x + 2y = x2
+ y2
– 2xy + 4
 x2 + y2 – 2xy – 2x – 2y + 4 = 0
h = –1, a = 1, b = 1, g = –1, f = –1, c = 4
  0 and h2
= ab, so curve is a parabola
Illustration 7 :
3x2  4y 2  48
Illustration 6 :

Two Standard Forms of the Ellipse
Standard Equation
x2

y 2
 
1 (a b)
a2
b2
(Horizontal Form of an Ellipse)
x2

y 2


1 (a, b)
b2
a2
(Vertical Form of an Ellipse)
Shape of the Ellipse
Centre (0, 0) (0, 0)
Equation of major axis y = 0 x = 0
Equation of minor axis x = 0 y = 0
Length of major axis 2a 2a
Length of minor axis 2b 2b
Foci (± ae, 0) (0, ± ae)
Vertices (± a, 0) (0, ± a)
Equation of directrices x  
a
e
y  
a
e
Eccentricity e 
a2  b2
e  a2  b2
a2
a2
Length of latus rectum
2b2
a
2b2
a
Ends of latra-recta

 ae, 


b2 

a 





b2
a

,  ae 


Parametric coordinates (a cos, b sin) (a cos, b sin)
Focal radii SP = a – ex1 and S′ P = a + ex1 SP = a – ey1 and S′ P = a + ey1
Sum of focal radii SP + S′P = 2a 2a
Distance between foci 2ae 2ae
Distance between directrices
2a
e
2a
e
Tangents at the vertices x = ± a y = ± a





2
CHORDS AND TANGENTS
Corresponding to any point P(x1, y1) we define the following expressions for the conic
S  ax2  2hxy  by 2  2gx  2fy  c  0
viz., T  axx1  h(xy1  x1y)  byy1  g (x  x1)  f (y  y1)  c
S1  ax2  2hx y  by2  2gx  2fy  c
1 1 1 1 1 1
x2

y 2


 
(i) The line y = mx + c intersects the ellipse
a2
two distinct points if c 2
 a2
m2
 b2
two coincident points if c 2
 a2
m2
 b2
imaginary points if c2
 a2
m2
 b2
1 at
b2
x2

y 2


(ii) Length of the chord intercepted by the ellipse
a2 1 on the line y = mx + c is
b2
2ab 1 m2 a2m2  b2  c 2
a2m2  b2
x2

y 2


(iii) Equation of tangent at the point (x1, y1) on the ellipse
a2
1 is
b2
T  0 ; i.e., xx1

yy1
 1
a2
b2
(iv) Equation of tangent at (a cos , b sin ) is x cos 

y sin 
 1
a b
(v) Equation of tangent in slope form is y  mx 

  a2m b2 
The point of contact of y  mx  a2m2  b2
is
 , 

a2m2  b2
 a2
m
a2m2  b2 


 b2 
and point of contact of y  mx  a2m2  b2
is
 , 


(vi) Point of intersection of two tangents

a2m2  b2 a2m2  b2 


a cos
 1  2 
b sin
 1  2 
  

2
 
2

Tangents at (a cos  , b sin  ) and (a cos  , b sin  ) intersect at   ,  
1 1 2 2
cos
 1  2 
cos
 1  2 


(vii) Equation of pair of tangents
   
   2 


x2

y 2


 
Through any given point P(x1, y1) there pass, in general, two tangents to the ellipse
a2
b2
1.
When P is external to the ellipse, the joint equation of the pair of tangents from P is SS1  T 2
a2m2  b2

11
16 cos2   11sin2 

Find the value of  if the tangent at the point 4 cos ,


sin 


to the ellipse 16x2 11y2  256 is also
a tangent to the circle x2  y 2  2x  15 .
Solution :
16x2 11y 2  256
x2

y 2

1
42
 
2

16

 …(1)
 


Equation of tangent at

P 4 cos ,


sin 


to (1) is
(4 cos )x  ( sin )y  16 …(2)
If (2) is a tangent to the circle x2  y2  2x 15  0 , then, length of the perpendicular from the center (1, 0)
to the line (2) is equal to the radius, i.e. 4 units
4 cos  16
  4
(cos   4)2  16 cos2  11sin2   5 cos2   11
 4 cos2   8 cos   5  0
(2 cos  1) (2 cos   5)  0
cos  
1
;
2
   


3
Chord of Contact
The equation of chord of contact of tangents drawn from P(x1, y1) is T  0
Director Circle
It is the locus of point of intersection of tangents which are perpendicular to each other. Its equation is
x2  y 2  a2  b2
Equation of Chord in Mid Point Form
Equation of the chord with middle point P(x1, y1) is T  S1
Illustration 8 :
16
11
16
11
11

a  b
 n 
2
n
 
Normals to the Ellipse
In general, four normals can be drawn from any point to an ellipse and the sum of the eccentric angles of
their feet is equal to an odd multiple of two right angles
(i) Equation of normal at P(x , y ) is x  x1

y  y1
1 1 x1 y1
a2
b2
(ii) Equation of normal at (a cos , b sin ) is ax sec   by cosec   a2  b2
m (b2
 a2
)
(iii) Equation of normal in slope form is y  mx 
b2
m2
 a2
Equation of chord joining two points
The equation of chord joining the points (a cos , b sin ) and (a cos , b sin ) is
x
cos
    

y
sin
    
 cos
    
 
a  2  b
 
 2 
 
 2 




Show that the straight line lx + my = n, is a normal to the ellipse x2

y 2
 1 if a2

b2  2 2 
2
   .
Solution :
a2
b2
l 2
m2  
x2

y 2


Equation of normal at any point P(a cos , b sin ) to the ellipse 1
a2 b
…(1)
is ax sec   by cosec   a2  b2
a sec 
 


 b cosec 
 
…(2)

a2  b2
If lx + my = n is also a normal to (1), we must have
l m n
 cos  
a

n
l a2  b2
; sin  
b

m
n
a2  b2
 a 2  b 2  a2  b2 
2
    


   


 l   m   

HYPERBOLA
The hyperbola is the locus of a point which moves such that its distance from a fixed point is e (> 1) times
its distance from a fixed straight line. As in case of an ellipse, the hyperbola has
(i) two foci
(ii) two directrices – one corresponding to each focus.
(iii) a center – the point such that all chords through it are bisected there at.
(iv) two axes – the transverse axis and conjugate axis.
(v) two vertices
(vi) two latus recta
Illustration 9 :

y
B M
P
S A Z C Z A S
B
e e
Note : The difference of the focal distances of any point on the hyperbola is equal to length of
transverse axis ; thus, PS  PS  2a .
Illustration 10 :
Thus, the hyperbola shown in the adjacent figure has the equation
x2

y 2

 1
a2
b2
(i) Centre is C(0, 0)
(ii) Vertices are A´(–a, 0) and A(a, 0)
(iii) Foci are S(ae, 0) ; S´(–ae, 0) x
(iv) Directrices are x  
a
e
(v) AA´ is the transverse axis and has length
2a ; BB´ is the conjugate axes where B
and B´ are two points on the axis of y
equidistant f rom C such that
CB = B´C = b and where
b2  a2(e2 1) .
x= –
a
x=
a
2b2
(vi) Length of latus rectum is
a
.
Find the equation of hyperbola with axes as co-ordinate axes and one of its vertex is at a distance 9 and 1
from two foci.
Solution :
PS – PS = 2a
 9 – 1 = 2a
 a = 4
More over focal distance is a + ae = 9
 4 + 4e = 9
5
e =
4
Since, b2
= a2
(e2
– 1)
 b2
 16
25  16 

 16 


= 9
b = 3
x2

y 2


 Equation of hyperbola is
16 9
1
Auxillary Circle
The circle described on the transverse axis AA´ of a hyperbola as diameter is called as auxillary circle of the
hyperbola and its equation is x2  y 2  a2 .

2
2 a


1 1 1 1

Equation of Hyperbola in Parametric Form
x = a sec  ; y = b tan , where 0    2 ( is the parameter)
x2

y 2


Position of a Point with respect to Hyperbola 1
a2
b2
(x)2

(y)2


 
The point R(x´, y´) lies within, upon or outside the hyperbola according as
a2
or negative.
Equilateral or Rectangular Hyperbola
1 is positive, zero
b2
The hyperbola x2
– y2
= a2
, whose asymptotes are at right angles (y = ± x) is called as an equilateral or
rectangular hyperbola. If this hyperbola is rotated such that the coordinate axes coincides with the asymptotes,
2
then, its equation reduces to xy = c2
where c  .
2
(i) Parametric form of xy = c2 is x = ct;
(ii) Centre is (0, 0)
(iii) Transverse axis has equation y = x
Conjugate axis has equation y = –x
(iv) Its eccentricity is e  .
y
y 
c
t
x
xy = c2
Tangent at

ct,

c 
t
 is t
2y  x  2ct
Normal at

ct,

c 
 is


xt3  ty ct4  c  0
The eccentricity of hyperbola 9x2
– 16y2
+ 72x – 32y – 16 = 0 is
Solution :
(x  4)2

(y 1)2


16 9
1, modified form of given equation
a2 = 16, b2 = 9
 2 b2 9 25 5
e  1
a2
 1 
16 16
so e 
4
Chords and Tangents
Corresponding to any point P(x1, y1), we define the following expressions for the conic
S  ax2  2hxy  by 2  2gx  2fy  c  0
viz., T  axx1  h(xy1  yx1)  byy1  g (x  x1)  f (y  y1)  c
S1  ax2  2hx y  by2  2gx  2fy  c
Illustration 11 :
1 1
t






  
x2

y 2


(i) The line y = mx + c intersects the hyperbola
a2
real and distinct point if c 2  a2m2  b2
real and coincident points if c2
 a2
m2
 b2
imaginary points if c2
 a2
m2
 b2
1 at
b2
x2

y 2


(ii) Length of the chord intercepted by the hyperbola
a2
1 on the line y = mx + c is
b2
x2

y 2


(iii) Equation of tangent at the point (x1, y1) lying on the hyperbola
a2
1 is T  0 ;
b2
i.e., xx1

yy1
 1
a2
b2
(iv) Equation of tangent at (a sec , b tan ) is x sec 

y tan 
 1
a b
(v) Equation of tangent in slope form is y  mx 

  a2m  b2 
The point of contact of y  mx  a2m2  b2
is
 , 

a2m2  b2 a2m2  b2 

 a2
m b2 
The point of contact of y  mx  a2m2  b2
is
 , 

Point of Intersection of Two Tangents

a2m2  b2 a2m2  b2 

The tangents at (a sec , b tan ) and (a sec , b tan ) intersect at
 a sin (2  1) b (cos 1  cos 2 )
 
 sin 2
,
 sin 1 sin 2

 sin 1 
Equation to Pair of Tangents
x2

y 2


 
Through any given point P(x1, y1) there pass in general, two tangents to the hyperbola
a2
1. When P
b2
is external to the hyperbola, the joint equation to the pair of tangents from P is SS1  T 2
 x2
y2   x2
y 2   xx yy 2
i.e.,   1  1
 1 1


  1  1 1
 a2
b2   2   a2 b2



Find the locus of the mid points of chords of the circle
9x2 16y 2  144 .
Solution :
x2  y2  16 which are tangents to the hyperbola
Let P( ) be any point on the locus ; then, P is the middle point of a chord of the circle
x2  y 2  16 …(1)
Illustration 12 :
2ab b2  c2  a2m2 1 m2
a2m2  b2
a2m2  b2
a2
b

y  y  
b x
 2

16



and is also a tangent to the hyperbola 9x2 16y2  144
x2

y 2

 1
42
32
…(2)
Equation of the chord is x   y   2  2 …(3)
If (3) is a tangent at (h, k) to (2) then (3) is same as the equation
9hx 16ky  144 …(4)



9h

16k

2  2


144
 h 
6 


2  2 ; k  
9 
2  2
h2

k 2


 




 2 2 2 2 2
Since (h, k) lies on (2) we have :
16 9
1; i.e., 16  9  (   )
Equation of locus is : 16x2  9y 2  (x2  y 2 )2
Chord of Contact
x2

y 2


 
Equation of chord of contact of the point (x1, y1) with respect to the hyperbola
a2
Director Circle
1 is T  0
b2
It is the locus of point of intersection of tangents which are perpendicular to each other. Its equation is
x2  y 2  a2  b2
Equation of Chord in Mid Point Form
Equation of the chord with middle point (x1, y1) is T  S1
Equation of Normal
(i) Equation of normal to the hyperbola
x2

y 2
a2
b2
 1 at the point (x1, y1) is
a2 y1
1 2
1
(x  x1)
(ii) Equation of normal at (a sec , b tan ) is a x cos   b y cot   a2  b2
SOLVED EXAMPLES
Example 1 :
Find the vertex, axis, focus, directrix and length of latus rectum of the parabola 2y2 + 3y – 4x – 3 = 0
Solution :
Given parabola is 2y2
+ 3y – 4x – 3 = 0
or 2y 

3
y

 4x  3  0
2 


or 2y 


3 2
 
4 
9 

 4x  3  0

t


 3 2 9  3 2 33
or 2y  
4   4x  3  0
8 or 2y  
4  4x  8
 
 3 2 
 
33 
or y  
4  2x  
32
   
This is of the form Y2
= 4aX where Y  y 
3
, 4a = 2, X =
4
x 
33
32
Equation of Axis is Y = 0  y 
3
 0
4
 y  
3
4
Vertex is (0,0) i.e. X = 0, Y = 0  x 
33
,
32
y  
3
4


33
, 
3 
 Vertex  
 32 4 
Focus is (a, 0) i.e. X = a, Y = 0
 x 
33

1
, y 
3
 0
32 2 4
 x  
17
,
32
y  
3
4
 17
,
 3 
 Focus is  
 32 4 
Directrix is X = –a
 x 
33
 
1
32 2
 x  
49
(Directrix)
32
Length of latus rectum = 4a = 2
Example 2 :
Find the equation of the parabola if the focus is at (6, –6) and vertex is (–2, 2).
Solution :
Let S and A be the focus and vertex of the parabola. Let axis
of parabola meet the directrix at M ( ); then, A is mid
point of line segment SM. Hence,
 + 6
–2 =
2
 − 6
2 =
2
  = –10
  = 10
Thus coordinates of M are (–10, 10).
Slope of directrix =
1


slope of axis 
1
2  6 

 2  6


 
The equation of directrix is given by
y – 10 = x + 10  x – y + 20 = 0 ...(1)

(x  6)2  (y  6)2
B
t1
(0, 0)
A
C
t2
1 2
1
 
Let P ( x, y) be any point on the parabola. Then 


 x2
– 12x + 36 + y2
+ 12y + 36 =
x2  y 2  400  40x  40y  2xy
2
 x2
+ y2
+ 2xy – 64x + 64y – 256 = 0
Example 3 :
Find the length of the side of an equilateral triangle inscribed in the parabola y2
= 4ax so that one angular
point is at the vertex.
Solution :
Let ABC be the equilateral triangle inscribed in the parabola y2 = 4ax
where A(0, 0), B (at 2, 2at1) and C (at 2, 2at2 ) . Since BC is
perpendicular to the axis of parabola, we have : t2 = –t1.
AB = BC  AB2 = BC2 X
 a2t 4  4a2t 2  16a2t 2
1 1 1
 t 4  12t2
1 1
 t 2  12
 Length of a side = 16a2
12  8 3a
Example 4 :
The normal at any point P on y2
= 4ax meets the axis in G and the tangent at the vertex in G´. If A be the vertex
and the rectangle AGQG´ be completed, show that the equation to the locus of Q is x3  2ax2  y2
Solution :
The normal at P(at2
, 2at) is
y + tx = 2at + at3 …(1)
The coordinates of G and G´ are
G(a(2  t 2 ), 0), G(0, at(2  t2 ))
Since AGQG´ is a rectangle, the co-ordinates (h, k) of Q are
x
h  a(2  t 2 ) and k  at (2  t2 )
  k 2 
 k = t h and h  a 2    

  h  

 h3  2ah2  ak 2
Equation of locus is : x3  2ax2  ay 2
Example 5 :
Through the vertex O of the parabola y2
= 4ax, chords OP and OQ are drawn at right angles to one another.
Show that for all positions of P, PQ cuts the axis of the parabola at a fixed point. Also find the locus of
the middle point of PQ.
x  y  20
2
y
G Q
P
A G

y
P
O
Q
i
Solution :
Let P and Q have coordinates (at2, 2at ) and (at 2, 2at ) respectively
1 1 2 2
Since OP  OQ , we have 
t1t2 = – 4 …(1)
Equation of PQ is
(t1  t2 )y  2x  8a
(x  4a) 
(t1  t2 )
y  0
x
2
This passes through the point (4a, 0) for all positions of P.
Let R(h, k) be any point on the locus ; then, R is the middle
point of PQ for some position of P.
a(t2  t2 )
 h  1 2 ; k  a(t1  t2 ) . Also t1t2  4
2
2h  k 2
  (t1  t2 )2  2t1t2     8
a  a 

k 2  2ah  8a2  0
Equation of locus is : y 2  2ax  8a2  0
Example 6 :
Show that the locus of the points such that two of the three normals to the parabola y2
= 4ax from them
coincide is 27ay 2  4 (x  2a)2
Solution :
Let P(h, k) be any point on the locus. The general equation of normal to y2
= 4ax is
y = mx – 2am – am3 …(1)
Let the three normals from P have slopes m1, m2 and m3 resp., where m2 = m3 ; then,
k  mi h  2ami  am3 : i 1, 2, 3
 m1, m2 and m3 are roots of the equation k  mh  2am  am3
 am3  (2a  h)m  k  0
 m1  m2  m3  0 …(2)
m1(m2  m3 )  m2m3 
2a  h
a
…(3)
m1 m2 m3
 
k
a
…(4)
Since m2 = m3, we have :
m1  2m2
m2

h  2a
(by(2)) …(5)
(by (3) and (5)) …(6)
2
3a

1
3
(x  6)2  (y  7)2
2
m3 
k
2a
(by (4) and (5)) …(7)
 h  2a 3  k 2
By (6) and (7) we have     
3a 2a
   
 Equation of locus is : 4 (x  2a)3  27ay 2
Example 7 :
Find the centre, the length of the axes, and the eccentricity of the ellipse 2x2  3y 2  4x  12y 13  0 .
Solution :
2x2  3y 2  4x 12y  13 = 0
2(x2  2x)  3(y 2  4y) 13 = 0
2(x 1)2  3(y  2)2 = 1
(x 1)2

(y  2)2
 1

2
  1
= 1

3

   

X 2

Y 2
a2
b2
= 1 ; where X =
Centre is : X = 0, Y = 0 ;
Length of major axes
Length of minor axes
 2a 

 2b 

If e denotes the eccentricity, then, b2  a2(1 e2 )

1

1
(1 e2 ) ; e 
3 2
Example 8 :
Find the equation of the ellipse whose focus is the point (6, 7) ; directrix is x + y + 2 = 0 and eccentricity
is .
Solution :
By definition of an ellipse, the focal distance of any point P(x, y) on it is ‘e’ times its distance from the directrix,
where ‘e’ denotes the eccentricity.
 

6 (x  6)2  6 (y  7)2
 (x  y  2)2
5x2  5y 2  76x  88y  2xy  506 = 0
x – 1, Y = y + 2 , a 
1
,
2
b 
1
3
x = 1, y = –2
2
2
3
1
3
1 x  y  2
3 2

5
b2 cos2   a2 sin2 
1
 d
1

2
Example 9 :
Find the equation of the ellipse whose foci are (2, 3), (–2, 3) and whose semi-minor axis is of length .
Solution :
Let S1(2, 3) and S2(–2, be the two foci and let 2a and 2b denote the lengths of major and minor axes
respectively. Then, b 
 ae = 2
b2  a2(1 e2 )
 a = 3
and 2ae = S1S2 = 4, where e is the eccentricity of the ellipse.
The major axes is y = 3 and center is (0, 3) – the mid point of the foci. Hence, equation of the ellipse is
x2

(y  3)2

1
9 5
Example 10 :
A tangent to the ellipse x2  4y 2  4 meets the ellipse x2  2y 2  6 at P and Q. Show that the tangents
at P and Q on the ellipse x2  2y 2  6 are perpendicular.
Solution :
Let A(2 cos , sin ) be any point on the ellipse x2  4y 2  4
Equation of tangent at A to (1) is x cos   2 sin   y  2
(2) intersects the ellipse x2  2y 2  6
…(1)
…(2)
…(3)
at P and Q ; Let T(h, k) be the point of intersection of the tangents at P and Q to (3) ; then equation of pair of
tangents from T to (3) is (x2  2y 2  6) (h2  2k 2  6)  (xh  2ky  6)2
x2(k 2  3)  y 2(h2  6)  2hk xy  6h x 12ky  3 (h2  2k 2 )  0
We need to show that h2  k 2  9
Equation of chord of contact of tangents from T to (3) is xh + 2ky = 6 …(4)
Since (2) and (4) represent the same line, we have,
cos 


h

sin 
k

3
; or h2  k 2  9
Let ‘d’ denote the perpendicular distance from the center of the ellipse x2

y 2


 
 to the tangent drawn
1
a2
b2
2 2  b2 

at a point P on the ellipse. If F1 and F2 are the two foci of the ellipse, show that (PF1  PF2 )
Solution :
 4a  

2 

x2

y 2


Let P(a cos , b sin ) be the point on the ellipse 1
a2 b
…(1)
The tangent at which is at a distance ‘d’ from the center of the ellipse. Equation of tangent at P to (1) is
b x cos   a y sin   ab
 d 
ab
…(2)
3)
5
Example 11 :

b
 0
d

b2

b2 cos2   a2 sin2 
d2
a2

b2

d2
(a2  b2 ) cos2 


a2
 e2 cos2 
where e denotes the eccentricity of the ellipse. For any point (x0, y0) on (1), the focal distances from foci F1 and
F2 are a + ex0 and a – ex0 respectively. Hence, (PF1  PF2 )2  (2e acos)2  4a2 e2 cos2 

 (PF1  PF2 )2  4a  2 
1 



 
 

Example 12 :
The tangent and normal to the ellipse x2  4y 2  4 at a point P on it meet the major axis in Q and R
respectively. If QR = 2, find the eccentric angle of P.
Solution :
x2  4y 2  4 ; a = 2, b = 1
The tangent at P(a cos , b sin ) is
ax cos   4by sin   4
x cos   2y sin   2
Equation of normal at P is
ax sec   by cosec   a2  b2
2x sin   y cos   3 sin  cos 

Coordinates of Q and R are given by : Q (2 sec , 0)R
 3
cos ,


 2 






…(1)
…(2)
QR = 2 
2 sec  
3
cos   2
2
 4  3 cos2    4 cos 
Taking ‘+’ sign, 3 cos2   4 cos   4  0
cos  
4  8
6
; cos  
2
3
Taking ‘–’ sign
3 cos2   4 cos   4  0
cos  

 cos   
2
3
Example 13 :
Find the eccentricity, foci, centre, directrices and the lengths of the transverse and conjugate axis of the
hyperbola x2  2y 2  2x  8y 1 0 .
2
4  64
6
2
1



Solution :
x2  2y 2  2x  8y 1 = 0
(x 1)2  2(y  2)2  6 = 0
(y  2)2

(x 1)2
3
Y 2

X 2
6
= 1
A2
B2 = 1 where Y = y – 2 and X = x – 1 ; A  and B 
The center is : X = 0, Y = 0 ; i.e. x = 1, y = 2.
Transverse axis is Y axes and conjugate axes is X axes.
 Length of transverse axis  2
Length of conjugate axis  2
If e denotes the eccentricity, then, B2  A2(e2 1)
e2
– 1 = 2; e 
Coordinates of foci are X = 0, Y = ± Ae
i.e. x = 1 and y  2  3  3
i.e. (1, 5) and (1, –1)
Equation of directrices are Y  
A
e
i.e. y = 3 and y = 1
Example 14 :
The equation of one of the directrices of a hyperbola is 2x + y = 1 and the corresponding focus is (1, 2)
and eccentricity e 
focus.
. Find the equation of the hyperbola and the co-ordinates of the center and second
Solution :
Let S be the focus of (1, 2) and PM be the perpendicular distance of a point P(x, y) on the hyperbola from
the corresponding directrix. Then, PS = e·PM.
 (x 1)2  (y  2)2
 (2x  y 1)
2
 3
 5 

5 (x2  y 2  2x  4y  5)  3 (4x2  y 2  4xy  4x  2y 1)
7x2  2y 2 12xy  2x  14y  22 = 0
is the equation of the hyperbola.
Equation of the perpendicular SZ from S to the
directrix 2x + y – 1 = 0 is x – 2y + 3 = 0
3 6
3
6
3
3
S(1, 2)
Z A
A

3
5
( 
( 
 




1
,
7 
 Co-ordinates of Z are  .
 5 

Since the vertices A and A´ divide SZ internally and externally respectively in the ratio
are :
:1, their coordinates
 5  10  7 3 
A   , 
 3 1)5 ( 3 1)5 
  (5  3) 7 3 10 
A   , 
 3 1)5 ( 3 1)5 
   8
,
11
 
The centre C of the hyperbola is the mid point of AA´ ; therefore C has coordinates 
10 10
 .
If S0 denotes the coordinates of the second focus, then C is the mid point of SS0.


13
,
1 
 Coordinates of S are  
 5 5 


  

3
0

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