JEE Mathematics/ Lakshmikanta Satapathy/ Two Dimensional Geometry QA part 9/ JEE question on Normal to Parabola and Tangent to Circle solved with the related concepts

1. Physics Helpline
L K Satapathy
Normal to Parabola = Tangent to Circle
2D Geometry QA 9
L
R
r

2. Physics Helpline
L K Satapathy
2D Geometry QA 9
( ) 2 ( ) 2 ( ) 3 ( ) 3a b c d
Question : If the normals to the parabola drawn at the end points of its latus
rectum are tangents to the circle , then the value of is2 2 2
( 3) ( 2)x y r   
2
4y x
2
r
Answer :
22 4
dy dy
y
dx dx y
  
L
R
(3 ,- 2)
(1 ,- 2)
(1 , 2)
r
The situation is shown in the figure
Equation of the given parabola is 2
4 . . . (1)y x
 Coordinates of the end points of LR = ( 1 ,  2 )
Equation of circle :
Differentiating (1) , we get
2 2 2
( 3) ( 2)x y r   
 Coordinates of its center = (3 , – 2)

3. Physics Helpline
L K Satapathy
2D Geometry QA 9
Correct option = (b)
We use the equation
If (3) is tangent to circle , then its distance from center = radius of circle
23 2 3 2 2 [ ]
2
r Anr s     
 Equation of normal at L (1 , 2) : 2 1( 1) 3 0 . . . (3)y x x y       
1
2N
T
y
m
m
  
 Slope of tangent 2
Tm
y

 Slope of normal
 Slope of normal at L (1 , 2) = –1
1 1
2 2
ax by c
d
a b
 



4. Physics Helpline
L K Satapathy
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