JEE Mathematics/ Lakshmikanta Satapathy/ 2D Geometry QA 12/ JEE Advanced Question on Image of point with respect to line solved using the formula derived in QA 10

1. Physics Helpline
L K Satapathy
Image of parabola
2D Geometry QA 12
( , )P  
1 1( , )x y
4 0x y  
O
5y  
Q
B A
C

2. Physics Helpline
L K Satapathy
2D Geometry QA 12
Question : Let the curve C be the mirror image of the parabola with
respect to the line x + y + 4 = 0 . If A and B be the points of intersection of C
with the line y = – 5 , then the distance between A and B is
( ) 2 ( ) 4 ( ) 6 ( ) 8a b c d
Answer :
We have derived the formula for coordinates of the
image point Q with respect to the line ax + by + c = 0
(in 2D Geometry QA 10) as follows :
2
4y x
Equation of give Line : 4 0 . . . (1)x y  
2
4y xLet P( , ) be a point on the parabola
And Q(x1 , y1) be its image w.r.to line (1)
( , )P  
1 1( , )x y
4 0x y  
O
5y  
Q
B A
C
2
4y x
1 1
2 2
2( )x y a b c
a b a b
       
 


3. Physics Helpline
L K Satapathy
2D Geometry QA 12
 The formula modifies to 1 1 2( 4)
1 1 1 1
x y       
 

1 1 ( 4)x y          
1 14 & 4x y       
 The coordinates of Q are given by 1 1( , ) ( 4 , 4)x y      
1 14 & 4x y       
The point (,) is on the parabola 2
4  
2
1 1 18 16 4 16x x y     
2
1 1( 4) 4( 4)x y     

4. Physics Helpline
L K Satapathy
2D Geometry QA 12
 EQN of C is given by 2
8 4 32 0 . . . (2)x x y   
2
( 4) 4( 4)x y     [ Parabola opening downward ]
2
1 1 18 4 32 0x x y    
Point of intersection with the line y = – 5
2
(2) 8 20 32 0x x     
( 2)( 6) 0x x   
2
8 12 0x x   
( 2, 5) & ( 6, 5)A B     
2 6x or x    
 The points of intersection are
Correct option = (b)
2 ( 6) [4 ]AnAB s     

5. Physics Helpline
L K Satapathy
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