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 Page 1


Solved Examples on Ellipse 
Q1: The length of the latus-rectum of the ellipse ?? ?? ?? + ?? ?? ?? = ???? is 
(a) v ?? /?? 
(b) v ?? /?? 
(c) ?? /?? 
(d) ???? /?? 
Ans: (d) 
Sol: Here the ellipse is 
?? 2
9
+
?? 2
5
= 1 
Here ?? 2
= 9 and ?? 2
= 5. So, latus-rectum =
2 ?? 2
?? =
2 ( 5 )
3
=
10
3
. 
Q2: What is the equation of the ellipse with foci ( ± ?? , ?? ) and eccentricity 
?? ?? 
(a) ?? ?? ?? + ?? ?? ?? = ???? 
(b) ?? ?? ?? + ?? ?? ?? = ???? 
(c) ?? ?? ?? + ?? ?? ?? = ?? 
(d) ?? ?? ?? + ?? ?? ?? = ?? 
Ans: (a) 
Sol: Here ???? = ± 2 , ? ?? =
1
2
, ? ?? = ± 4 
Form ?? 2
= ?? 2
( 1 - ?? 2
) ? ?? 2
= 16 ( 1 -
1
4
) ? ?? 2
= 12 
Hence, the equation of ellipse is 
?? 2
16
+
?? 2
12
= 1 or 3 ?? 2
+ 4 ?? 2
= 48 
Q3: The eccentricity of an ellipse, with its centre at the origin is 
?? ?? . If one of 
the directrices is ?? = ?? , then the equation of the ellipse is 
(a) ?? ?? ?? + ?? ?? ?? = ?? 
(b) ?? ?? ?? + ?? ?? ?? = ???? 
(c) ?? ?? ?? + ?? ?? ?? = ???? 
(d) ?? ?? ?? + ?? ?? ?? = ?? 
Ans: (b) 
Sol: Given ?? =
1
2
,
?? ?? = 4. So, ?? = 2 ? ?? 2
= 4 
Page 2


Solved Examples on Ellipse 
Q1: The length of the latus-rectum of the ellipse ?? ?? ?? + ?? ?? ?? = ???? is 
(a) v ?? /?? 
(b) v ?? /?? 
(c) ?? /?? 
(d) ???? /?? 
Ans: (d) 
Sol: Here the ellipse is 
?? 2
9
+
?? 2
5
= 1 
Here ?? 2
= 9 and ?? 2
= 5. So, latus-rectum =
2 ?? 2
?? =
2 ( 5 )
3
=
10
3
. 
Q2: What is the equation of the ellipse with foci ( ± ?? , ?? ) and eccentricity 
?? ?? 
(a) ?? ?? ?? + ?? ?? ?? = ???? 
(b) ?? ?? ?? + ?? ?? ?? = ???? 
(c) ?? ?? ?? + ?? ?? ?? = ?? 
(d) ?? ?? ?? + ?? ?? ?? = ?? 
Ans: (a) 
Sol: Here ???? = ± 2 , ? ?? =
1
2
, ? ?? = ± 4 
Form ?? 2
= ?? 2
( 1 - ?? 2
) ? ?? 2
= 16 ( 1 -
1
4
) ? ?? 2
= 12 
Hence, the equation of ellipse is 
?? 2
16
+
?? 2
12
= 1 or 3 ?? 2
+ 4 ?? 2
= 48 
Q3: The eccentricity of an ellipse, with its centre at the origin is 
?? ?? . If one of 
the directrices is ?? = ?? , then the equation of the ellipse is 
(a) ?? ?? ?? + ?? ?? ?? = ?? 
(b) ?? ?? ?? + ?? ?? ?? = ???? 
(c) ?? ?? ?? + ?? ?? ?? = ???? 
(d) ?? ?? ?? + ?? ?? ?? = ?? 
Ans: (b) 
Sol: Given ?? =
1
2
,
?? ?? = 4. So, ?? = 2 ? ?? 2
= 4 
From ?? 2
= ?? 2
( 1 - ?? 2
) ? ?? 2
= 4 ( 1 -
1
4
) = 4 ×
3
4
= 3 
Hence the equation of ellipse is 
?? 2
4
+
?? 2
3
= 1, i.e. 3 ?? 2
+ 4 ?? 2
= 12 
Q4: The curve represented by ?? = ?? ( ?? ?? ?? ? ?? + ?? ???? ? ?? ) , ?? = ?? ( ?? ?? ?? ? ?? - ?? ???? ? ?? ) is 
(a) Ellipse 
(b) Parabola 
(c) Hyperbola 
(d) Circle 
Ans: (a) 
Sol: Given, ?? = 3 ( cos ? ?? + sin ? ?? ) , ?? = 4 ( cos ? ?? - sin ? ?? ) ?
?? 3
= ( cos ? ?? + sin ? ?? ) ,
?? 4
=
( cos ? ?? - sin ? ?? ) 
Squaring and adding, we get 
?? 2
9
+
?? 2
16
= ( 1 + sin ? 2 ?? ) + ( 1 - sin ? 2 ?? ) ?
?? 2
9
+
?? 2
16
= 2, 
which represents ellipse. 
Q5: The foci of the ellipse ???? ( ?? + ?? )
?? + ?? ( ?? + ?? )
?? = ?????? are 
(a) ( - ?? , ?? ) , ( ?? , ?? ) 
(b) ( - ?? , - ?? ) , ( ?? , ?? ) 
(c) ( ?? , - ?? ) , ( ?? , - ?? ) 
(d) ( - ?? , ?? ) , ( - ?? , - ?? ) 
Ans: (d) 
Sol: Given ellipse is 
( ?? + 1 )
2
9
+
( ?? + 2 )
2
25
= 1 i.e. 
?? 2
9
+
?? 2
25
= 1, where ?? = ?? + 1 and ?? =
?? + 2 
Here ?? 2
= 25 , ?? 2
= 9 [Type :
?? 2
?? 2
+
?? 2
?? 2
= 1 ] 
Eccentricity is given by ?? 2
=
?? 2
- ?? 2
?? 2
=
25 - 9
25
=
16
25
, ? ?? =
4
5
 
Foci are given by ?? = ± ???? = ± 5 (
4
5
) = ± 4 
?? = 0 ? ?? + 2 = ± 4 ? ?? = - 2 ± 4 = - 6 or 2 
?? + 1 = 0 ? ?? = - 1. Hence foci are ( - 1 , - 6 ) or ( - 1 , 2 ). 
 
Page 3


Solved Examples on Ellipse 
Q1: The length of the latus-rectum of the ellipse ?? ?? ?? + ?? ?? ?? = ???? is 
(a) v ?? /?? 
(b) v ?? /?? 
(c) ?? /?? 
(d) ???? /?? 
Ans: (d) 
Sol: Here the ellipse is 
?? 2
9
+
?? 2
5
= 1 
Here ?? 2
= 9 and ?? 2
= 5. So, latus-rectum =
2 ?? 2
?? =
2 ( 5 )
3
=
10
3
. 
Q2: What is the equation of the ellipse with foci ( ± ?? , ?? ) and eccentricity 
?? ?? 
(a) ?? ?? ?? + ?? ?? ?? = ???? 
(b) ?? ?? ?? + ?? ?? ?? = ???? 
(c) ?? ?? ?? + ?? ?? ?? = ?? 
(d) ?? ?? ?? + ?? ?? ?? = ?? 
Ans: (a) 
Sol: Here ???? = ± 2 , ? ?? =
1
2
, ? ?? = ± 4 
Form ?? 2
= ?? 2
( 1 - ?? 2
) ? ?? 2
= 16 ( 1 -
1
4
) ? ?? 2
= 12 
Hence, the equation of ellipse is 
?? 2
16
+
?? 2
12
= 1 or 3 ?? 2
+ 4 ?? 2
= 48 
Q3: The eccentricity of an ellipse, with its centre at the origin is 
?? ?? . If one of 
the directrices is ?? = ?? , then the equation of the ellipse is 
(a) ?? ?? ?? + ?? ?? ?? = ?? 
(b) ?? ?? ?? + ?? ?? ?? = ???? 
(c) ?? ?? ?? + ?? ?? ?? = ???? 
(d) ?? ?? ?? + ?? ?? ?? = ?? 
Ans: (b) 
Sol: Given ?? =
1
2
,
?? ?? = 4. So, ?? = 2 ? ?? 2
= 4 
From ?? 2
= ?? 2
( 1 - ?? 2
) ? ?? 2
= 4 ( 1 -
1
4
) = 4 ×
3
4
= 3 
Hence the equation of ellipse is 
?? 2
4
+
?? 2
3
= 1, i.e. 3 ?? 2
+ 4 ?? 2
= 12 
Q4: The curve represented by ?? = ?? ( ?? ?? ?? ? ?? + ?? ???? ? ?? ) , ?? = ?? ( ?? ?? ?? ? ?? - ?? ???? ? ?? ) is 
(a) Ellipse 
(b) Parabola 
(c) Hyperbola 
(d) Circle 
Ans: (a) 
Sol: Given, ?? = 3 ( cos ? ?? + sin ? ?? ) , ?? = 4 ( cos ? ?? - sin ? ?? ) ?
?? 3
= ( cos ? ?? + sin ? ?? ) ,
?? 4
=
( cos ? ?? - sin ? ?? ) 
Squaring and adding, we get 
?? 2
9
+
?? 2
16
= ( 1 + sin ? 2 ?? ) + ( 1 - sin ? 2 ?? ) ?
?? 2
9
+
?? 2
16
= 2, 
which represents ellipse. 
Q5: The foci of the ellipse ???? ( ?? + ?? )
?? + ?? ( ?? + ?? )
?? = ?????? are 
(a) ( - ?? , ?? ) , ( ?? , ?? ) 
(b) ( - ?? , - ?? ) , ( ?? , ?? ) 
(c) ( ?? , - ?? ) , ( ?? , - ?? ) 
(d) ( - ?? , ?? ) , ( - ?? , - ?? ) 
Ans: (d) 
Sol: Given ellipse is 
( ?? + 1 )
2
9
+
( ?? + 2 )
2
25
= 1 i.e. 
?? 2
9
+
?? 2
25
= 1, where ?? = ?? + 1 and ?? =
?? + 2 
Here ?? 2
= 25 , ?? 2
= 9 [Type :
?? 2
?? 2
+
?? 2
?? 2
= 1 ] 
Eccentricity is given by ?? 2
=
?? 2
- ?? 2
?? 2
=
25 - 9
25
=
16
25
, ? ?? =
4
5
 
Foci are given by ?? = ± ???? = ± 5 (
4
5
) = ± 4 
?? = 0 ? ?? + 2 = ± 4 ? ?? = - 2 ± 4 = - 6 or 2 
?? + 1 = 0 ? ?? = - 1. Hence foci are ( - 1 , - 6 ) or ( - 1 , 2 ). 
 
Q6: If any tangent to the ellipse 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? intercepts equal lengths ?? on the 
axes, then ?? = 
(a) ?? ?? + ?? ?? 
(b) v ?? ?? + ?? ?? 
(c) ( ?? ?? + ?? ?? )
?? 
(d) None of these 
Ans: (b) 
Sol: The equation of any tangent to the given ellipse is 
?? ?? cos ? ?? +
?? ?? sin ? ?? = 1 
This line meets the coordinate axes at ?? (
?? cos ? ?? , 0 ) and ?? (0 ,
?? s in ? ?? ) 
?
?? cos ? ?? = ?? =
?? s in ? ?? ? cos ? ?? =
?? ?? and sin ? ?? =
?? ?? ? cos
2
? ?? + sin
2
? ?? =
?? 2
?? 2
+
?? 2
?? 2
? ?? 2
= ?? 2
+
?? 2
? ?? = v ?? 2
+ ?? 2
. 
Q7: The locus of the middle point of the intercept of the tangents drawn 
from an external point to the ellipse ?? ?? + ?? ?? ?? = ?? between the coordinate 
axes, is 
(a) 
?? ?? ?? +
?? ?? ?? ?? = ?? 
(b) 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? 
(c) 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? 
(d) 
?? ?? ?? ?? +
?? ?? ?? = ?? 
Ans: (c) 
Sol: Let the point of contact be ?? = ( v 2 cos ? ?? , sin ? ?? ) 
Equation of tangent ???? is 
?? v 2
cos ? ?? + ?? sin ? ?? = 1 
? ?? = ( v 2 sec ? ?? , 0 ) ; ?? = ( 0 , coses ? ?? ) 
Let the middle point ?? of ???? be ( h , ?? ). 
? h =
sec ? ?? v 2
, ?? =
cosec ? ?? 2
? cos ? ?? =
1
h v 2
, sin ? ?? =
1
2 ?? ?
1
2 h
2
+
1
4 ?? 2
= 1 
Page 4


Solved Examples on Ellipse 
Q1: The length of the latus-rectum of the ellipse ?? ?? ?? + ?? ?? ?? = ???? is 
(a) v ?? /?? 
(b) v ?? /?? 
(c) ?? /?? 
(d) ???? /?? 
Ans: (d) 
Sol: Here the ellipse is 
?? 2
9
+
?? 2
5
= 1 
Here ?? 2
= 9 and ?? 2
= 5. So, latus-rectum =
2 ?? 2
?? =
2 ( 5 )
3
=
10
3
. 
Q2: What is the equation of the ellipse with foci ( ± ?? , ?? ) and eccentricity 
?? ?? 
(a) ?? ?? ?? + ?? ?? ?? = ???? 
(b) ?? ?? ?? + ?? ?? ?? = ???? 
(c) ?? ?? ?? + ?? ?? ?? = ?? 
(d) ?? ?? ?? + ?? ?? ?? = ?? 
Ans: (a) 
Sol: Here ???? = ± 2 , ? ?? =
1
2
, ? ?? = ± 4 
Form ?? 2
= ?? 2
( 1 - ?? 2
) ? ?? 2
= 16 ( 1 -
1
4
) ? ?? 2
= 12 
Hence, the equation of ellipse is 
?? 2
16
+
?? 2
12
= 1 or 3 ?? 2
+ 4 ?? 2
= 48 
Q3: The eccentricity of an ellipse, with its centre at the origin is 
?? ?? . If one of 
the directrices is ?? = ?? , then the equation of the ellipse is 
(a) ?? ?? ?? + ?? ?? ?? = ?? 
(b) ?? ?? ?? + ?? ?? ?? = ???? 
(c) ?? ?? ?? + ?? ?? ?? = ???? 
(d) ?? ?? ?? + ?? ?? ?? = ?? 
Ans: (b) 
Sol: Given ?? =
1
2
,
?? ?? = 4. So, ?? = 2 ? ?? 2
= 4 
From ?? 2
= ?? 2
( 1 - ?? 2
) ? ?? 2
= 4 ( 1 -
1
4
) = 4 ×
3
4
= 3 
Hence the equation of ellipse is 
?? 2
4
+
?? 2
3
= 1, i.e. 3 ?? 2
+ 4 ?? 2
= 12 
Q4: The curve represented by ?? = ?? ( ?? ?? ?? ? ?? + ?? ???? ? ?? ) , ?? = ?? ( ?? ?? ?? ? ?? - ?? ???? ? ?? ) is 
(a) Ellipse 
(b) Parabola 
(c) Hyperbola 
(d) Circle 
Ans: (a) 
Sol: Given, ?? = 3 ( cos ? ?? + sin ? ?? ) , ?? = 4 ( cos ? ?? - sin ? ?? ) ?
?? 3
= ( cos ? ?? + sin ? ?? ) ,
?? 4
=
( cos ? ?? - sin ? ?? ) 
Squaring and adding, we get 
?? 2
9
+
?? 2
16
= ( 1 + sin ? 2 ?? ) + ( 1 - sin ? 2 ?? ) ?
?? 2
9
+
?? 2
16
= 2, 
which represents ellipse. 
Q5: The foci of the ellipse ???? ( ?? + ?? )
?? + ?? ( ?? + ?? )
?? = ?????? are 
(a) ( - ?? , ?? ) , ( ?? , ?? ) 
(b) ( - ?? , - ?? ) , ( ?? , ?? ) 
(c) ( ?? , - ?? ) , ( ?? , - ?? ) 
(d) ( - ?? , ?? ) , ( - ?? , - ?? ) 
Ans: (d) 
Sol: Given ellipse is 
( ?? + 1 )
2
9
+
( ?? + 2 )
2
25
= 1 i.e. 
?? 2
9
+
?? 2
25
= 1, where ?? = ?? + 1 and ?? =
?? + 2 
Here ?? 2
= 25 , ?? 2
= 9 [Type :
?? 2
?? 2
+
?? 2
?? 2
= 1 ] 
Eccentricity is given by ?? 2
=
?? 2
- ?? 2
?? 2
=
25 - 9
25
=
16
25
, ? ?? =
4
5
 
Foci are given by ?? = ± ???? = ± 5 (
4
5
) = ± 4 
?? = 0 ? ?? + 2 = ± 4 ? ?? = - 2 ± 4 = - 6 or 2 
?? + 1 = 0 ? ?? = - 1. Hence foci are ( - 1 , - 6 ) or ( - 1 , 2 ). 
 
Q6: If any tangent to the ellipse 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? intercepts equal lengths ?? on the 
axes, then ?? = 
(a) ?? ?? + ?? ?? 
(b) v ?? ?? + ?? ?? 
(c) ( ?? ?? + ?? ?? )
?? 
(d) None of these 
Ans: (b) 
Sol: The equation of any tangent to the given ellipse is 
?? ?? cos ? ?? +
?? ?? sin ? ?? = 1 
This line meets the coordinate axes at ?? (
?? cos ? ?? , 0 ) and ?? (0 ,
?? s in ? ?? ) 
?
?? cos ? ?? = ?? =
?? s in ? ?? ? cos ? ?? =
?? ?? and sin ? ?? =
?? ?? ? cos
2
? ?? + sin
2
? ?? =
?? 2
?? 2
+
?? 2
?? 2
? ?? 2
= ?? 2
+
?? 2
? ?? = v ?? 2
+ ?? 2
. 
Q7: The locus of the middle point of the intercept of the tangents drawn 
from an external point to the ellipse ?? ?? + ?? ?? ?? = ?? between the coordinate 
axes, is 
(a) 
?? ?? ?? +
?? ?? ?? ?? = ?? 
(b) 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? 
(c) 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? 
(d) 
?? ?? ?? ?? +
?? ?? ?? = ?? 
Ans: (c) 
Sol: Let the point of contact be ?? = ( v 2 cos ? ?? , sin ? ?? ) 
Equation of tangent ???? is 
?? v 2
cos ? ?? + ?? sin ? ?? = 1 
? ?? = ( v 2 sec ? ?? , 0 ) ; ?? = ( 0 , coses ? ?? ) 
Let the middle point ?? of ???? be ( h , ?? ). 
? h =
sec ? ?? v 2
, ?? =
cosec ? ?? 2
? cos ? ?? =
1
h v 2
, sin ? ?? =
1
2 ?? ?
1
2 h
2
+
1
4 ?? 2
= 1 
Thus required locus is 
1
2 ?? 2
+
1
4 ?? 2
= 1 
Q8: Find the centre, the eccentricity, the foci, the directrices and the 
lengths and the equations of the axes of the ellipse 5x
2
 + 9y
2
 + 10x - 36y - 
4= 0  
Sol: Rewrite the equation in the standard form and compare them to get the 
centre, eccentricity etc.  
5x
2
 + 9y
2
 + 10x - 36y - 4= 0 , the given equation can be written as  
5(x
2
 + 2x) + 9(y
2
 - 4y) = 4  
5(x + 1)
2
 + 9(y - 2)
2
 = 45 
 
Shift the origin to O ' = ( -1, 2) 
? X = x+1; Y = y-2 
  ……(i)  
This is in standard form 
 
Now for an ellipse in the standard form we have Centre = (0, 0); foci = (±ae, 0); 
directrices  axes x = 0, y = 0, length of major axis = 2a, length of minor 
axis = 2b. 
Now for (i) the centre is given by X = 0, Y = 0  
? x + 1= 0, y - 2 = 0 
 
Page 5


Solved Examples on Ellipse 
Q1: The length of the latus-rectum of the ellipse ?? ?? ?? + ?? ?? ?? = ???? is 
(a) v ?? /?? 
(b) v ?? /?? 
(c) ?? /?? 
(d) ???? /?? 
Ans: (d) 
Sol: Here the ellipse is 
?? 2
9
+
?? 2
5
= 1 
Here ?? 2
= 9 and ?? 2
= 5. So, latus-rectum =
2 ?? 2
?? =
2 ( 5 )
3
=
10
3
. 
Q2: What is the equation of the ellipse with foci ( ± ?? , ?? ) and eccentricity 
?? ?? 
(a) ?? ?? ?? + ?? ?? ?? = ???? 
(b) ?? ?? ?? + ?? ?? ?? = ???? 
(c) ?? ?? ?? + ?? ?? ?? = ?? 
(d) ?? ?? ?? + ?? ?? ?? = ?? 
Ans: (a) 
Sol: Here ???? = ± 2 , ? ?? =
1
2
, ? ?? = ± 4 
Form ?? 2
= ?? 2
( 1 - ?? 2
) ? ?? 2
= 16 ( 1 -
1
4
) ? ?? 2
= 12 
Hence, the equation of ellipse is 
?? 2
16
+
?? 2
12
= 1 or 3 ?? 2
+ 4 ?? 2
= 48 
Q3: The eccentricity of an ellipse, with its centre at the origin is 
?? ?? . If one of 
the directrices is ?? = ?? , then the equation of the ellipse is 
(a) ?? ?? ?? + ?? ?? ?? = ?? 
(b) ?? ?? ?? + ?? ?? ?? = ???? 
(c) ?? ?? ?? + ?? ?? ?? = ???? 
(d) ?? ?? ?? + ?? ?? ?? = ?? 
Ans: (b) 
Sol: Given ?? =
1
2
,
?? ?? = 4. So, ?? = 2 ? ?? 2
= 4 
From ?? 2
= ?? 2
( 1 - ?? 2
) ? ?? 2
= 4 ( 1 -
1
4
) = 4 ×
3
4
= 3 
Hence the equation of ellipse is 
?? 2
4
+
?? 2
3
= 1, i.e. 3 ?? 2
+ 4 ?? 2
= 12 
Q4: The curve represented by ?? = ?? ( ?? ?? ?? ? ?? + ?? ???? ? ?? ) , ?? = ?? ( ?? ?? ?? ? ?? - ?? ???? ? ?? ) is 
(a) Ellipse 
(b) Parabola 
(c) Hyperbola 
(d) Circle 
Ans: (a) 
Sol: Given, ?? = 3 ( cos ? ?? + sin ? ?? ) , ?? = 4 ( cos ? ?? - sin ? ?? ) ?
?? 3
= ( cos ? ?? + sin ? ?? ) ,
?? 4
=
( cos ? ?? - sin ? ?? ) 
Squaring and adding, we get 
?? 2
9
+
?? 2
16
= ( 1 + sin ? 2 ?? ) + ( 1 - sin ? 2 ?? ) ?
?? 2
9
+
?? 2
16
= 2, 
which represents ellipse. 
Q5: The foci of the ellipse ???? ( ?? + ?? )
?? + ?? ( ?? + ?? )
?? = ?????? are 
(a) ( - ?? , ?? ) , ( ?? , ?? ) 
(b) ( - ?? , - ?? ) , ( ?? , ?? ) 
(c) ( ?? , - ?? ) , ( ?? , - ?? ) 
(d) ( - ?? , ?? ) , ( - ?? , - ?? ) 
Ans: (d) 
Sol: Given ellipse is 
( ?? + 1 )
2
9
+
( ?? + 2 )
2
25
= 1 i.e. 
?? 2
9
+
?? 2
25
= 1, where ?? = ?? + 1 and ?? =
?? + 2 
Here ?? 2
= 25 , ?? 2
= 9 [Type :
?? 2
?? 2
+
?? 2
?? 2
= 1 ] 
Eccentricity is given by ?? 2
=
?? 2
- ?? 2
?? 2
=
25 - 9
25
=
16
25
, ? ?? =
4
5
 
Foci are given by ?? = ± ???? = ± 5 (
4
5
) = ± 4 
?? = 0 ? ?? + 2 = ± 4 ? ?? = - 2 ± 4 = - 6 or 2 
?? + 1 = 0 ? ?? = - 1. Hence foci are ( - 1 , - 6 ) or ( - 1 , 2 ). 
 
Q6: If any tangent to the ellipse 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? intercepts equal lengths ?? on the 
axes, then ?? = 
(a) ?? ?? + ?? ?? 
(b) v ?? ?? + ?? ?? 
(c) ( ?? ?? + ?? ?? )
?? 
(d) None of these 
Ans: (b) 
Sol: The equation of any tangent to the given ellipse is 
?? ?? cos ? ?? +
?? ?? sin ? ?? = 1 
This line meets the coordinate axes at ?? (
?? cos ? ?? , 0 ) and ?? (0 ,
?? s in ? ?? ) 
?
?? cos ? ?? = ?? =
?? s in ? ?? ? cos ? ?? =
?? ?? and sin ? ?? =
?? ?? ? cos
2
? ?? + sin
2
? ?? =
?? 2
?? 2
+
?? 2
?? 2
? ?? 2
= ?? 2
+
?? 2
? ?? = v ?? 2
+ ?? 2
. 
Q7: The locus of the middle point of the intercept of the tangents drawn 
from an external point to the ellipse ?? ?? + ?? ?? ?? = ?? between the coordinate 
axes, is 
(a) 
?? ?? ?? +
?? ?? ?? ?? = ?? 
(b) 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? 
(c) 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? 
(d) 
?? ?? ?? ?? +
?? ?? ?? = ?? 
Ans: (c) 
Sol: Let the point of contact be ?? = ( v 2 cos ? ?? , sin ? ?? ) 
Equation of tangent ???? is 
?? v 2
cos ? ?? + ?? sin ? ?? = 1 
? ?? = ( v 2 sec ? ?? , 0 ) ; ?? = ( 0 , coses ? ?? ) 
Let the middle point ?? of ???? be ( h , ?? ). 
? h =
sec ? ?? v 2
, ?? =
cosec ? ?? 2
? cos ? ?? =
1
h v 2
, sin ? ?? =
1
2 ?? ?
1
2 h
2
+
1
4 ?? 2
= 1 
Thus required locus is 
1
2 ?? 2
+
1
4 ?? 2
= 1 
Q8: Find the centre, the eccentricity, the foci, the directrices and the 
lengths and the equations of the axes of the ellipse 5x
2
 + 9y
2
 + 10x - 36y - 
4= 0  
Sol: Rewrite the equation in the standard form and compare them to get the 
centre, eccentricity etc.  
5x
2
 + 9y
2
 + 10x - 36y - 4= 0 , the given equation can be written as  
5(x
2
 + 2x) + 9(y
2
 - 4y) = 4  
5(x + 1)
2
 + 9(y - 2)
2
 = 45 
 
Shift the origin to O ' = ( -1, 2) 
? X = x+1; Y = y-2 
  ……(i)  
This is in standard form 
 
Now for an ellipse in the standard form we have Centre = (0, 0); foci = (±ae, 0); 
directrices  axes x = 0, y = 0, length of major axis = 2a, length of minor 
axis = 2b. 
Now for (i) the centre is given by X = 0, Y = 0  
? x + 1= 0, y - 2 = 0 
 
Foci are given by X = ±ae, Y= 0 
i.e. x +1 = ±2 and y - 2 = 0  
i.e. x = 1, y = 2 and x = -3, y= 2  
? Foci = (1, 2); (-3, 2) 
The equation of directrices are given by  
 
The equation of the axes are given by  
X = 0, Y = 0 
i.e. x + 1 = 0, y - 2 = 0  
i.e. x = -1, y = 2  
Length of the axes being 2a, 2b 
i.e, 6, 2v5. 
Q9: A tangent to the ellipse  touches it at the point P in the first 
quadrant and meets the x and y axes in A and B respectively. If P divides 
AB in the ratio 3 : 1, find the equation of the tangent at P. 
Sol: Consider a point in the parametric form and obtain the points A and B. Now 
use the condition that the point P divides AB in the ratio 3:1. 
Let P = (a cos ?, b sin ?): 
   …….(i)  
Equation of the tangent at P(?) is
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Document Description: Solved Examples: Ellipse for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The notes and questions for Solved Examples: Ellipse have been prepared according to the JEE exam syllabus. Information about Solved Examples: Ellipse covers topics like and Solved Examples: Ellipse Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Solved Examples: Ellipse.
Introduction of Solved Examples: Ellipse in English is available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & Solved Examples: Ellipse in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: Solved Examples: Ellipse | Mathematics (Maths) for JEE Main & Advanced
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