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Physics Helpline
L K Satapathy
JEE Question on Ellipse
2D Geometry QA 5
Physics Helpline
L K Satapathy
2D Geometry QA 5
Question : The ellipse is inscribed in a rectangle R whose
sides are parallel to the coordinate axes. Another ellipse , passing through the
point (0,4) circumscribes the rectangle R . Then the eccentricity of the ellipse is
22
1 : 1
9 4
yxE  
2E
2E
31 1 3( ) ( ) ( ) ( )
2 2 42
a b c d
Answer :
 Ellipse is bounded by the lines x =  3 and y =  2
 Equations of sides of rectangle ABCD are x =  3 and y =  2
Vertices are A (– 3 , 2 ) B ( 3 , 2) C ( 3 , – 2 ) D ( – 3 , – 2 )
Equation of given ellipse :
22
1 . . . (1)
9 4
yx  
A
D
B
C
x
y
O
2E
1E
2 2
9 4a and b  
Physics Helpline
L K Satapathy
2D Geometry QA 5
2
2 2
9 4 9 3(3,2) (3) 1 12
16 4
B a
a a
       
2 2 2
16 12 4c a b     
2 1
4 2
ce
a
   
22
1
12 16
yx  
2 2
16 , 12a b  
Correct option = (c)
2
2 2 2
0 16(0,4) 1 16E b
a b
     
Let the equation of ellipse E2 be
22
2 2
1 . . . (2)
yx
a b
 
22
2
(2) 1 . . . (3)
9
yx
a
   
 The equation of ellipse E2 is
Physics Helpline
L K Satapathy
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2D Geometry QA 5

  • 1. Physics Helpline L K Satapathy JEE Question on Ellipse 2D Geometry QA 5
  • 2. Physics Helpline L K Satapathy 2D Geometry QA 5 Question : The ellipse is inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse , passing through the point (0,4) circumscribes the rectangle R . Then the eccentricity of the ellipse is 22 1 : 1 9 4 yxE   2E 2E 31 1 3( ) ( ) ( ) ( ) 2 2 42 a b c d Answer :  Ellipse is bounded by the lines x =  3 and y =  2  Equations of sides of rectangle ABCD are x =  3 and y =  2 Vertices are A (– 3 , 2 ) B ( 3 , 2) C ( 3 , – 2 ) D ( – 3 , – 2 ) Equation of given ellipse : 22 1 . . . (1) 9 4 yx   A D B C x y O 2E 1E 2 2 9 4a and b  
  • 3. Physics Helpline L K Satapathy 2D Geometry QA 5 2 2 2 9 4 9 3(3,2) (3) 1 12 16 4 B a a a         2 2 2 16 12 4c a b      2 1 4 2 ce a     22 1 12 16 yx   2 2 16 , 12a b   Correct option = (c) 2 2 2 2 0 16(0,4) 1 16E b a b       Let the equation of ellipse E2 be 22 2 2 1 . . . (2) yx a b   22 2 (2) 1 . . . (3) 9 yx a      The equation of ellipse E2 is
  • 4. Physics Helpline L K Satapathy For More details: www.physics-helpline.com Subscribe our channel: youtube.com/physics-helpline Follow us on Facebook and Twitter: facebook.com/physics-helpline twitter.com/physics-helpline