(5) Ellipse (Theory). Module-3pdf

Total no.of questions in Ellipse are-
Inchapter examples................................................. 11
Solved Examples..................................................... 22
Total No. of questions............................................ 33
ELLIPSE

1 DEFINITION
An ellipse is the locus of a point which moves in
such a way that its distance form a fixed point is in
constant ratio to its distance from a fixed line. The
fixed point is called the focus and fixed line is called
the directrix and the constant ratio is called the
eccentricity of a ellipse denoted by (e).
In other word, we can say an ellipse is the locus of
a point which moves in a plane so that the sum of
it distances from fixed points is constant.
2. EQUATION OFAN ELLIPSE
2.1 Standard Form of the equation of ellipse
Let the distance between two fixed points S and S' be
2ae and let C be the mid point of SS.
Taking CS as x- axis, C as origin.
Let P(h,k) be the moving point Let SP+ SP = 2a
(fixed distance) then
SP+S'P= }
k
)
ae
h
{( 2
2

 + }
k
)
ae
h
{( 2
2

 = 2a
h2
(1– e2
) + k2
= a2
(1– e2
)
Hence Locus of P(h, k) is given by.
x2
(1– e2
) + y2
= a2
(1– e2
)
 2
2
a
x
+
)
e
1
(
a
y
2
2
2

= 1
Let us assume that a2
(1– e2
)= b2
 The standard equation will be given by
1
b
y
a
x
2
2
2
2


2.1.1 Various parameter related with standard
ellipse :
Let the equation of the ellipse 1
b
y
a
x
2
2
2
2

 a > b
(i) Vertices of an ellipse : The point of which
ellipse cut the axis x-axis at (a,0) & (– a, 0)
and y- axis at (0, b) & (0, – b) is called the
vertices of an ellipse.
(ii) Major & Minor axis : The straight line AA
is called major axis and BB is called minor
axis. The major and minor axis taken
together are called the principal axes and its
length will be given by
Length of major axis  2a
Length of minor axis  2b
(iii) Centre : The point which bisect each chord
of an ellipse is called centre (0,0) denoted by 'C'.
(iv) Directrix : ZM and Z M are two directrix and
their equation are x= a/e and x = – a/e.
(v) Focus : S (ae, 0) and S (–ae,0) are two foci
of an ellipse.
(vi) Latus Rectum : Such chord which passes
through either focus and perpendicular to the
major axis is called its latus rectum.
Length of Latus Rectum :
If L is (ae,) then 2 is the length of
Latus Rectum.
Length of Latus rectum is given by
a
b
2 2
.
(vii) Relation between constant a, b, and e
b2
= a2
(1– e2
)  e2
= 2
2
2
a
b
a 
 e = 2
2
2
a
b
a 
Result :
(a) Centre C is the point of intersection of the
axes of an ellipse. Also C is the mid point of AA.
(b) Another form of standard equation of ellipse
1
b
y
a
x
2
2
2
2

 when a < b.
In this case major axis is BB= 2b which is
along y- axis and minor axis is AA= 2a along
x- axis. Focus S(0,be) and S(0,–be) and
directrix are y = b/e and y = –b/e.
2.2 General equation of the ellipse
The general equation of an ellipse whose focus is
(h,k) and the directrix is the line ax + by + c = 0
and the eccentricity will be e. Then let P(x1
,y1
) be any
point on the ellipse which moves such that SP = ePM
 (x1
–h)2
+ (y1
–k)2
= 2
2
2
1
1
2
b
a
)
c
by
ax
(
e



Hence the locus of (x1
,y1
) will be given by
(a2
+ b2
) [(x – h)2
+ (y–k)2
] = e2
(ax + by + c)2
Which is the equation of second degree from which
we can say that any equation of second degree
represent equation of an ellipse.
Minor Axis
Major Axis
Directrix
x = -a/e
Directrix
x = a/e
(–ae, 0) (ae, 0)
S’
S
A A’

•
•
S(1,1)
P(x,y)
x
-
y
+3
=
0
M
Note : Condition for second degree in X & Y to
represent an ellipse is that if h2
= ab < 0 &
= abc + 2 fgh – af2
– bg2
– ch2
 0
Examples
based on Equation of an ellipse
Ex.1Find the equation of the ellipse having its
M
M
Q
A C
S S
centre at the point (2,–3), one focus at
(3,– 3) and one vertex at (4,–3).
Sol. C  (2, – 3), S  (3, – 3) and A  (4, – 3)
Now CA = 2
2
)
3
3
(
)
2
4
( 


 = 2  a = 2
Again CS = 2
2
)
3
3
(
)
2
3
( 


 = 1
 ae = 1  e =
a
1
=
2
1
Let the directrix cut the major- axis at Q.
Then
AQ
AS
= e =
2
1
If Q  (, ), then
SA : AQ = e : 1 = 1 : 2
A  




 



3
6
,
3
6
= (4, – 3)

3
6


= 4,
3
6


= – 3
  = 6,  = – 3
Slope of CA = 0, therefore directrix will be
parallel to y- axis.
Since directrix is parallel to y- axis and it
passes through Q(6, – 3)
 equation of the directrix is x = 6.
Let P(x,y) be any point on the ellipse, then
e =
2
1
=
PM
PS
=
   
2
2
2
1
6
x
3
y
3
x





4
1
=
   
 2
2
2
6
x
3
y
3
x




;
or (x – 6)2
= 4 [(x– 3)2
+ (y + 3)2
]
or x2
–12x+36 = 4 [x2
– 6x + 9 +y2
+ 6y + 9]
or 3x2
+ 4y2
– 12x + 24y + 36 = 0
Ex.2 Find the centre, the length of the axes,
eccentricity and the foci of the ellipse.
12x2
+ 4y2
+ 24 x – 16 y + 25 = 0
Sol. The given equation can be written in the form
12(x + 1)2
+ 4(y – 2)2
= 3

 
4
/
1
1
x 2

+
 
4
/
3
2
y
2

= 1 ...(1)
Co- ordinates of centre of the ellipse are given
by x + 1 = 0 and y – 2 = 0
Hence centre of the ellipse is (– 1, 2)
If a and b be the lengths of the semi major
and semi-minor axes, then a2
=3/4, b2
= 1/4
 Length of major axis = 2a =3,
Length of minor axis = 2b = 1
 a =
2
3
, b =
2
1
Since b2
= a2
(1 – e2
) 1/4 = 3/4 (1– e2
)
 e = 3
/
2  ae =
2
3
×
3
2
=
2
1
Co-ordinates of foci are given by
x+1 = 0, y–2 = ± ae
Thus foci are 









2
1
2
,
1
Ex.3 Find the equation of an ellipse whose focus is
(– 1,1), eccentricity is 1/2 and the directrix is
x – y + 3 = 0.
Sol. Let P (x,y) be
any point on the ellipse whose
focus is S (–1,1) and the directrix is
x – y + 3 = 0.
PM
perpendicular
from P (x,y) on
the directrix
x –y + 3 = 0.
Then by
definition
SP = ePM
(SP)2
=e2
(PM)2

 (x + 1)2
+(y – 1)2
=
4
1
2
2
3
y
x





 

8 (x2
+ y2
+ 2x –2y+ 2)
= x2
+ y2
+ 9 – 2xy + 6x – 6y
 7x2
+7y2
+ 2xy+10x–10y +7 = 0
which is the required equation of the ellipse.
Ex.4 The foci of an ellipse are (± 2, 0) and its
eccentricity is 1/2, find its equation.
Sol. Let the equation of the ellipse be 2
2
a
x
+ 2
2
b
y
= 1,
Then coordinates of foci are (± ae, 0).
 ae = 2  a×
2
1
= 2 






2
1
e

 a = 4
We have b2
= a2
(1– e2
)
 b2
=16 






4
1
1 =12
Thus, the equation of the ellipse is
12
y
16
x 2
2
 = 1
3. Parametric form of the ellipse
Let the equation of ellipse in standard form will be
given by 2
2
a
x
+ 2
2
b
y
= 1
Then the equation of ellipse in the parametric form
will be given by x= a cos , y = b sin  where  is
the eccentric angle whose value vary from 0   < 2.
Therefore coordinate of any point P on the ellipse
will be given by (a cos, b sin ).
4. POINT AND ELLIPSE
Let P(x1
,y1
) be any point and let 2
2
a
x
+ 2
2
b
y
= 1 is the
equation of an ellipse.
The point lies outside, on or inside the ellipse as if
S1
= 2
2
1
a
x
+ 2
2
1
b
y
– 1 > 0, = 0, < 0
Ex.5 Find the position of the point (4, – 3) relative
to the ellipse 5x2
+ 7y2
= 140.
Sol. 5(4)2
+ 7(– 3)2
–140 = 80 + 63 –140 =3 > 0,
so the point (4,–3) lies outside the ellipse
5x2
+ 7y2
= 140.
5. ELLIPSE AND A LINE
Let the ellipse be 2
2
a
x
+ 2
2
b
y
= 1 and the given line
be y = mx + c.
Solving the line and ellipse we get
2
2
a
x
+ 2
2
b
)
c
mx
( 
= 1
i.e. (a2
m2
+ b2
) x2
+ 2 mca2
x + a2
(c2
– b2
) = 0
above equation being a quadratic in x.
 discriminant = 4m2
c2
a4
–4a2
(a2
m2
+ b2
)(c2
– b2
)
= –b2
{c2
–(a2
m2
+ b2
)}
= b2
{(a2
m2
+b2
)– c2
}
Hence the line intersects the parabolas in
2 distinct points if a2
m2
+b2
> c2
, in one point if c2
=
a2
m2
+b2
, and does not intersect if a2
m2
+b2
< c2
.
 y = mx ±  
2
2
2
b
m
a  touches the ellipse and
condition for tangency c2
= a2
m2
+ b2
.
Hence y = mx ±  
2
2
2
b
m
a  , touches the ellipse
2
2
a
x
+ 2
2
b
y
=1at 











2
2
2
2
2
2
2
2
b
m
a
b
,
b
m
a
m
a
.
Note : 1: x cos + y sin = p is a tangent if
p2
= a2
cos2
+ b2
sin2
.
Note : 2 : x + my + n = 0 is a tangent if
n2
= a2
2
+ b2
m2
.
5.1 Equation of the Tangent :
(i) The equation of the tangent at any point
(x1
, y1
) on the ellipse 2
2
a
x
+ 2
2
b
y
=1 is
2
1
a
xx
+ 2
1
b
yy
= 1.
(ii) the equation of tangent at any point ‘’ is
a
x
cos  +
b
y
sin  = 1.
5.2 Equation of the Normal :
(i) The equation of the normal at any point
(x1
, y1
) on the ellipse 2
2
a
x
+ 2
2
b
y
= 1 is
1
2
x
x
a
–
1
2
y
y
b
= a2
– b2
(ii) The equation of the normal at any point ‘’ is
ax sec  – by cosec  = a2
– b2

Examples
based on Ellipse and a line
Ex.6Find the condition that the line x + my = n
may be a normal to the ellipse 2
2
a
x
+ 2
2
b
y
= 1.
Sol. equation of normal to the given ellipse at
(a cos, b sin) is

cos
ax
–

sin
by
= a2
– b2
...(1)
If the line x + my = n is also normal to the
ellipse then there must be a value of  for
which line (1) and line x + my = n are
identical. For that value of  we have







cos
a

=








sin
b
m
=
 
2
2
b
a
n

or cos =
 
2
2
b
a
.
an


...(2)
and sin =
 
2
2
b
a
.
m
bn


...(3)
Squarring and adding (2) and (3), we get
1 =
 2
2
2
2
b
a
n









 2
2
2
2
m
b
a

which is the required condition.
Ex.7For what value of  does the line y = x + 
touches the ellipse 9x2
+ 16y2
= 144.
Sol.  Equation of ellipse is
9x2
+ 16y2
= 144 or
16
x2
+
9
y2
= 1
Comparing this with 2
2
a
x
+ 2
2
b
y
= 1
then we get a2
= 16 and b2
= 9
& comparing the line y = x + with y = mx +c
 m = 1 and c = 
If the line y = x +  touches the ellipse
9x2
+ 16y2
= 144, then
c2
= a2
m2
+ b2

= 16 × 12
+ 9
2
= 25
  = ± 5
Ex.8 Find the equations of the tangents to the ellipse
3x2
+ 4y2
= 12 which are perpendicular to the
line y + 2x = 4.
Sol. Let m be the slope of the tangent, since the
tangent is perpendicular to the line y + 2x = 4.
 m × –2 = –1  m =
2
1
Since 3x2
+ 4y2
= 12
or
4
x2
+
3
y2
= 1
Comparing this with 2
2
a
x
+ 2
2
b
y
= 1
 a2
= 4 and b2
= 3
So the equation of the tangents are
y =
2
1
x ± 3
4
1
x
4 
 y = 2
x
2
1
 or x – 2y ± 4 = 0
Ex.9 Find the equations of tangent to the ellipse
4x2
+ 9y2
= 36 at the point (3, –2).
Sol. We have 4x2
+ 9y2
= 36

9
x2
+
4
y2
= 1. This is of the form
2
2
a
x
+ 2
2
b
y
= 1, where a2
= 9 and b2
= 4.
We know that the equation of the tangent to
the ellipse 2
2
a
x
+ 2
2
b
y
= 1 at (x1
,y1
) is
2
a
xx1
+ 2
b
yy1
= 1
So, the equation of the tangent to the given
ellipse at (3,– 2) is
9
x
3
–
4
y
2
= 1 i.e.
3
x
–
2
y
= 1
Ex.10 Find the equations of the tangents to the
ellipse x2
+ 16y2
= 16 each one of which
makes an angle of 60º with the x- axis.
Sol. We have, x2
+ 16y2
= 16  2
2
4
x
+ 2
2
1
y
= 1
This is of the form 2
2
a
x
+ 2
2
b
y
= 1, where
a2
= 16 and b2
= 1
So, the equations of the tangents are
y = mx ± 2
2
2
b
m
a 
i.e. y = 3 x ± 1
3
x
16 
 y = 3 x ± 7

Ex.11 If the normal at an end of a latus- rectum of
an ellipse 2
2
a
x
+ 2
2
b
y
= 1 passes through one
extermity of the minor axis, show that the
eccentricity of the ellipse is given by
e4
+ e2
– 1 = 0
or e2
=
2
1
5 
Sol. The co-ordinates of an end of the
latus- rectum are (ae, b2
/a). The equation of
normal at P(ae,b2
/a) is
ae
x
a2
–
a
/
b
)
y
(
b
2
2
= a2
– b2
or
e
ax
– ay = a2
– b2
It passes through one extremity of the minor
axis whose co-ordinates are (0, – b)
 0 + ab = a2
– b2
or (a2
b2
) = (a2
– b2
)2
or a2
.a2
(1– e2
) = (a2
e2
)2
or 1– e2
= e4
or e4
+ e2
– 1 = 0 or (e2
)2
+ e2
– 1 = 0
 e2
=
2
4
1
1 


e2
=
2
1
5 
(taking positive sign)
6. PAIR OF TANGENTS
Let P(x1
, y1
) be any point lies outside the ellipse
2
2
a
x
+ 2
2
b
y
= 1, and let a pair of tangents PA,PB can
be drawn to it from P.
then the equation of pair of tangents of PA and PB
is SS1
= T2
where S  2
2
a
x
+ 2
2
b
y
– 1 = 0
S1
 2
2
a
x
+ 2
2
b
y
– 1 = 0, T  2
1
a
xx
+ 2
1
b
yy
– 1 = 0
Note : The locus of the point of intersection of the
tangents to an ellipse 2
2
a
x
+ 2
2
b
y
= 1 which
are perpendicular to each other is called
Director circle.
Hence the equation of director circle of the
ellipse 2
2
a
x
+ 2
2
b
y
= 1 is x2
+y2
= a2
+b2
.
7. CHORD OF CONTACT
If PA and PB be the tangents through point
P(x1
, y1
) to the ellipse 2
2
a
x
+ 2
2
b
y
= 1,
then the equation of the chord of contact AB is
2
1
a
xx
+ 2
1
b
yy
= 1 or T = 0 (at x1
, y1
)
8. EQUATION OF CHORD WITH MID POINT
(x1
, y1
)
The equation of the chord of the ellipse
2
2
a
x
+ 2
2
b
y
= 1, whose mid point be (x1
,y1
) is
T = S1
Where T 
2
1
a
xx
+
2
1
b
yy
– 1=0,S1
 2
2
1
a
x
+ 2
2
1
b
y
– 1 = 0
9. DIAMETER
The locus of the middle points of a system of parallel
chords is called a diameter. If y = mx + c represent a
system of parallel chords of the ellipse 2
2
a
x
+ 2
2
b
y
= 1
then the line y = –
m
a
b
2
2
x is the equation of the
diameter.
A (a,0)
B’ (0,-b)
(0,b) B
(-a,0)A’ S
P 







a
b
,
ae
2

Ex.1 The equation of the ellipse which passes through
origin and has its foci at the points (1, 0) and (3,
0) is-
(A) 3x2
+ 4y2
= x (B) 3x2
+ y2
= 12x
(C) x2
+ 4y2
= 12x (D) 3x2
+ 4y2
= 12x
Sol:[D] Centre being mid point of the foci is





 
0
,
2
3
1
= (2, 0)
Distance between foci 2ae = 2
ae = 1 or a2
– b2
= 1 ...(i)
If the ellipse 2
2
a
)
2
x
( 
+ 2
2
b
y
= 1, then as it passes
from (0, 0)
2
a
4
= 1  a2
= 4
from (i) b2
= 3
Hence
4
)
2
x
( 2

+
3
y2
= 1
or 3x2
+ 4y2
– 12x = 0
Ex.2 A man running round a racecourse notes that the
sum of the distance of two flag posts from him is
always 10 meters and the distance between the
flag posts is 8 meters. The area of the path he
encloses -
(A) 10 (B) 15 (C) 5 (D) 20
Sol.:[B] The race course will be an ellipse with the flag
posts as its foci. If a and b are the semi major
and minor axes of the ellipse, then sum of focal
distances 2a = 10 and 2ae = 8
a = 5, e = 4/5
 b2
= a2
(1 – e2
) = 25 






25
16
1 = 9
Area of the ellipse = ab
=.5.3 =15
Ex.3 The distance of a point on the ellipse
6
x2
+
2
y2
= 1 from the centre is 2. Then eccentric
angle of the point is
(A) ±
2

(B) ± 
(C)
4

,
4
3
(D) ±
4

Sol:[C] Any point on the ellipse is
( 6 cos , 2 sin ), where  is an eccentric
angle.
It's distance from the center (0, 0) is given 
SOLVED EXAMPLES
6 cos2
 + 2 sin2
 = 4
or 3 cos2
 + sin2
 = 2
2 cos2
 = 1
 cos  = ±
2
1
;  =
4

,
4
3
Ex.4 If x cos  + y sin  = P is a tangent to the ellipse
2
2
a
x
+ 2
2
b
y
= 1, then -
(A) a cos  + b sin  = P2
(B) a sin  + b cos  = P2
(C) a2
cos2
 + b2
sin2
 = P2
(D) a2
sin2
 + b2
cos2
 = P2
Sol:[C] Given line is x cos  + y sin  = P ... (1)
Any tangent to the ellipse is
a
cos
x 
+
b
sin
y 
= 1 ...(2)
Comparing (1) and (2)


cos
a
cos
=


sin
b
sin
=
P
1
cos  =
P
cos
a 
; sin  =
P
sin
b 
Eliminate ,
cos2
 + sin2
 = 2
2
2
P
cos
a 
+ 2
2
2
P
sin
b 
or a2
cos2
 + b2
sin2
 = P2
Ex.5 The equation of tangents to the ellipse
9x2
+ 16y2
= 144 which pass through the point
(2, 3)-
(A) y = 3 (B) x + y = 2
(C) x – y = 3 (D) y = 3 ; x + y = 5
Sol.[D] Ellipse 9x2
+ 16y2
= 144
or
16
x2
+
9
y2
= 1
Any tangent is y = mx + 9
16m 2
 it passes
through (2, 3)
3 = 2m + 9
m
16 2

(3 – 2m)2
= 16m2
+ 9
m = 0, –1
Hence the tangents are y = 3, x + y = 5
Ex.6 If tan 1
tan 2
= – 2
2
b
a
, then the chord joining
two point 1
and 2
on the ellipse 2
2
a
x
+ 2
2
b
y
=1
will subtend a right angle at
(A) Focus (B) Centre
(C) End of the major axes (D) End of minor
axes

Sol.[B] Let P(a cos 1
, b sin 1
) and Q(a cos 2
, b sin 2
)
be two points on the ellipse. Then
m1
= Slope of OP (O is an centre) =
a
b
tan 1
;
m2
= Slope of OQ =
a
b
tan 2
 m1
m2
=
a
b
tan 
.
a
b
tan 
= 2
2
a
b
tan 1
tan 2
( tan 1
tan 2
= 2
2
b
a

)
= 2
2
a
b








 2
2
b
a
= – 1 POQ
 = 90°
Hence PQ makes a right angle at the centre of
the ellipse.
Ex.7 Chords of an ellipse are drawn through the
positive end of the minor axes. Then their mid
point lies on -
(A) a circle (B) a parabola
(C) an ellipse (D) a hyperbola
Sol.[C] Let (h, k) be the mid point of a chord passing
through the positive end of the minor axis of the
ellipse 2
2
a
x
+ 2
2
b
y
= 1. Then the equation of the
chord, T = S1
2
a
hx
+ 2
b
ky
– 1 = 2
2
a
h
+ 2
2
b
k
– 1
or 2
a
hx
+ 2
b
ky
= 2
2
a
h
+ 2
2
b
k
This passes through (0, b), therefore
b
k
= 2
2
a
h
+ 2
2
b
k
Hence, the locus of (h, k) is 2
2
a
x
+ 2
2
b
y
=
b
y
,
which is an ellipse.
Ex.8 The line x= at2
meets the ellipse 2
2
a
x
+ 2
2
b
y
= 1 in
the real points if -
(A) |t| < 2 (B) |t|  1
(C) |t| >1 (D) None of these
Sol.[B] Putting x = at2
in the equation of the ellipse, we
get
2
2
2
a
t
a
+ 2
2
b
y
= 1  y2
= b2
(1 – t4
)
y2
= b2
(1 – t2
) (1 + t2
)
This will give real values of y if
(1 – t2
)  0 |t|  1
Ex.9 The eccentric angles of the extremities of latus
rectum of the ellipse 2
2
a
x
+ 2
2
b
y
= 1 is-
(A) tan–1 




 
b
ae
(B) tan–1 




 
a
be
(C) tan–1 




 
ae
b
(D) tan–1 




 
be
a
Sol.[C] The coordinate of any point on the ellipse
2
2
a
x
+ 2
2
b
y
= 1 whose eccentric angle  are (a cos
, b sin ) .The coordinate of the end point of
latus rectum are 








a
b
,
ae
2
a cos  = ae and
b sin  =±
a
b2
tan =±
ae
b
  = tan–1







ae
b
Ex.10 The equation x2
+ 4y2
+ 2x + 16y + 13 = 0
represents a ellipse -
(A) whose eccentricity is 3
(B) whose focus is (± 3 , 0)
(C) whose directrix is x = ±
3
4
– 1
(D) None of these
Sol.[C] We have x2
+ 4y2
+ 2x + 16y + 13 = 0
(x2
+ 2x + 1) + 4(y2
+ 2y + 4) = 4
(x + 1)2
+ 4(y + 2)2
= 4
2
2
2
)
1
x
( 
+ 2
2
1
)
2
y
( 
= 1
Shifting the origin at (–1, –2) without rotating
the coordinate axes
2
2
2
x
+ 2
2
1
y
= 1 where x = X – 1
y = Y – 2
2
a
x
+ 2
2
b
y
=1,
where a = 2, b = 1
eccentricity of the ellipse
e = 2
2
a
b
1 =
4
1
1 =
2
3
Focus of the ellipse (± ae, 0)
X = x + 1 = ± 2.
2
3
 x = ± 3 – 1
Y = y + 2 = 0  y = – 2
Directrix of the ellipse X = ± a/e
x + 1 = ±
2
/
3
2
; x = ±
3
4
– 1

Ex.11 The line x + my + n = 0 cut the ellipse
2
2
a
x
+ 2
2
b
y
= 1 in points whose eccentric angle
differ by /2. Then the value of a2
2
+ b2
m2
is-
(A) 2n2
(B) 2n (C) 2m2
(D) 2m
Sol.[A] Suppose the line x + my + n = 0 cuts the ellipse
at P(a cos , b sin ) and
Q(a cos (/2 + ), b sin (/2 + )).
Then these two point lie on the line
a cos  + mb sin  + n = 0
– a sin  + mb cos  + n = 0
a cos  + mb sin  = – n ...(i)
– a sin  + mb cos  = – n ...(ii)
Square and add the equations (i) and (ii)
(a cos  + mb sin )2
+ (–a sin  + mb cos )2
= n2
+ n2
2
a2
(cos2
 + sin2
) +m2
b2
(sin2
 + cos2
)
= n2
+n2
2
a2
+ m2
b2
= 2n2
Ex.12 Product of the perpendiculars from the foci upon
any tangent to the ellipse 2
2
a
x
+ 2
2
b
y
= 1 is-
(A) b (B) a (C) a2
(D) b2
Sol.[D] The equation of any tangent to the ellipse
2
2
a
x
+ 2
2
b
y
= 1 is y = mx + 2
2
2
b
m
a 
 mx – y + 2
2
2
b
m
a  =0 ...(i)
The two foci of the given ellipse are S(ae, 0) and
S (–ae, 0). let p1
and p2
be the lengths of
perpendicular from S and S respectively on (i),
Then
p1
= length of perpendicular from S(ae, 0) on (i)
p1
=
1
m
b
m
a
mae
2
2
2
2



p2
= length of perpendicular from S(–ae, 0) on (i)
p2
=
1
m
b
m
a
mae
2
2
2
2




Now p1
p2
= 










1
m
b
m
a
mae
2
2
2
2












1
m
b
m
a
mae
2
2
2
2
= 2
2
2
2
2
m
1
b
)
e
1
(
m
a



 b2
= a2
(1 – e2
)
= 2
2
2
2
m
1
b
b
m


=
1
m
)
1
m
(
b
2
2
2


= b2
Ex.13 The point of the intersection of the tangent at
the two point on the ellipse 2
2
a
x
+ 2
2
b
y
= 1
whose eccentricity differ by a right angle lies on
the ellipse is-
(A) 2
2
a
x
+ 2
2
b
y
= 2 (B)
a
x
+
b
y
= 2
(C) 2
2
a
x
– 2
2
b
y
= 1 (D) 2
2
a
x
+ 2
2
b
y
= 1
Sol.[A] Let P(a cos , b sin ) and Q(a cos , b sin ) be
two points an the ellipse such that
 –  =
2

. The equation of tangent at
P and Q are respectively
a
x
cos  +
b
y
sin  = 1 ...(i)
and
a
x
cos  +
b
y
sin  = 1 ...(ii)
since  –  =
2

, so (i) can be written as
–
a
x
sin  +
b
y
cos  = 1 ...(iii)
Squaring (ii) and (iii) and then adding, we get
2
sin
b
y
cos
a
x








 +
2
cos
b
y
sin
a
x









 = 1 + 1
2
2
a
x
+ 2
2
b
y
= 2
Ex.14 Find the equation of the ellipse whose
eccentricity is 1/2, the focus is (–1, 1) and the
directrix is x – y + 3 = 0.
Sol. Let P (x,y) be any point on the ellipse whose
focus is S(–1,1) and eccentricity e =1/2.
Let PM be perpendicular from P on the directrix.
Then,
SP = ePM  SP =
2
1
(PM)  4 (SP)2
= PM2
 4 [(x +1)2
+ (y –1)2
] =
 
2
2
2
1
1
3
y
x














 8 (x2
+ y2
+ 2x – 2y + 2) = (x– y+3)2
 7x2
+ 7y2
+ 10x – 10y + 2xy + 7 = 0
This is the required equation of the ellipse.
Ex.15 Find the equation of the ellipse whose axes are
along the coordinate axes, vertices are (± 5,0)
and foci at (± 4,0).
Sol. Let the equation of the required ellipse be
2
2
a
x
+ 2
2
b
y
= 1 ...(1)

The coordinates of its vertices and foci are
(± a, 0) and (± ae,0) respectively.
 a = 5 and ae = 4  e = 4/5.
Now, b2
= a2
(1– e2
)  b2
= 25 






25
16
1 = 9.
Substituting the values of a2
and b2
in (1) , we
get
25
x2
+
9
y2
= 1,
which is the equation of the required ellipse.
Ex.16 Find the centre, the length of the axes and the
eccentricityof theellipse 2x2
+3y2
–4x–12y+13 = 0.
Sol. The given equation can be rewritten as
2[x2
– 2x] + 3 [y2
– 4y] + 13 = 0
or 2 (x – 1)2
+ 3 (y– 2)2
= 1
or 2
2
)
2
/
1
(
)
1
x
( 
+ 2
2
)
3
/
1
(
)
2
y
( 
= 1,
or 2
2
a
X
+ 2
2
b
Y
= 1
 Centre X = 0, Y = 0 i.e. (1,2).
Length of major axis = 2a = 2
Length of minor axis = 2b = 2/ 3 and
e = a
/
)
b
a
( 2
2
 = 1/ 3
Ex.17 P,Q,R be three points  on the ellipse x2
/
a2
+ y2
/b2
= 1. Show that the area of the triangle
formed by the tangents at P,Q.R is
ab tan
2
1
( – ) tan
2
1
( – ) tan
2
1
(– )
Sol. Equation of tangents at P() , Q() are
a
cos
x 
+
b
sin
y 
= 1 and
a
cos
x 
+
b
sin
y 
= 1,
These intersects in (x1
,y1
), then
x1
= a cos 1/2 (+ ) / cos 1/2 (– );
y1
= b sin 1/2 (+ ) / cos 1/2 ( – ).
Similarly, other points of intersections.
Ex.18 Find the equations of the tangents to the ellipse
4x2
+ 3y2
= 5 which are inclined at
an angle of 60º to the axis of x. Also, find the
point of contact.
Sol. The slope of the tangent = tan 60º = 3
Now, 4x2
+ 3y2
= 5 
4
/
5
x 2
+
3
/
5
y2
=1
2
a
x
+ 2
2
b
y
= 1, where
a2
=
4
5
and b2
=
3
5 .
We know that the equations
of the tangents of slope m to the ellipse
2
2
a
x
+ 2
2
b
y
= 1 are given by y = mx ± 2
2
2
b
m
a 
and the coordinates of the points of contact are











2
2
2
2
2
2
2
2
b
m
a
b
,
b
m
a
m
a

Here, m = 3 , a2
= 5/4 and b2
= 5/3.
So, the equations of the tangents are
y = 3 x ± 





3
4
5
+
3
5
i.e. y = 3 x ±
12
65
The coordinates of the points of contact are









12
/
65
3
/
5
,
12
/
65
4
/
3
5
 i.e 








39
195
2
,
26
65
3
 .
Ex.19 The radius of the circle passing through the foci of
the ellipse
16
x2
+
9
y2
= 1, and having its
centre (0,3) is-
(A) 4 (B) 3 (C) 12 (D) 7/2
Sol.[A] e = 1 – 2
2
a
b
= 1 –
16
9
 e =
4
7
 Foci are (± ae, 0) or (± 7 ,0).
Centre is (0,3)  Radius = 9
7  = 4
Ex.20 The eccentricity of the ellipse represented by the
equation 25x2
+ 16y2
– 150 x – 175 = 0 is-
(A) 2/5 (B) 3/5
(C) 4/5 (D) None of these
Sol.[B] 25(x2
– 6x + 9) + 16y2
= 175 + 225
or 25(x – 3)2
+16y2
=400 or
16
X2
+
25
Y2
= 1. Form
2
2
b
X
+ 2
2
a
Y
= 1
 Major axis lies along y- axis. ;
 e2
= 1 – 2
2
a
b
= 1 –
25
16
;  e =
5
3
Ex.21 The number of values of c such that the straight
line y = 4x + c touches the curve
4
x2
+ y2
= 1 is-
(A) 0 (B) 1
(C) 2 (D) infinite

Sol.[C] We know that the line y = mx + c touches the
curve 2
2
a
x
+ 2
2
b
y
= 1 if c2
= a2
m2
+ b2
Here, a2
= 4, b2
= 1, m = 4;
 c2
= 64 + 1  c = ± 65
Ex.22 If the chords of contact of tangents from two
points (x1
,y1
) and (x2
,y2
) to the ellipse
2
2
a
x
+ 2
2
b
y
= 1 are at right angles, then
2
1
2
1
y
y
x
x
is
equal to-
(A) 2
2
b
a
(B) – 2
2
a
b
(C) – 4
4
b
a
(D) – 4
4
a
b
Sol.[C] The equations of the chords of contact of tangents
drawn from (x1
,y1
) and (x2
,y2
) to the ellipse
2
2
a
x
+ 2
2
b
y
= 1 are 2
1
a
x
x
+ 2
1
b
y
y
= 1
... (i)
2
2
a
x
x
+ 2
2
b
y
y
= 1 ... (ii)
It is given that (i) and (ii) are at right angles.
 2
2
a
b

1
1
y
x
x 2
2
a
b

2
2
y
x
= – 1

2
1
2
1
y
y
x
x
= – 4
4
b
a

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