B
C
A
D
x-y = 1
x+y = 1
x+y = 3
x-y = 3
By Team Aakash Byju's | 14th December 2022
2
Let the normal at (h, k) on the curve x = 4y passes through (1,2). Then the slope of the tangent be
2
x = 4y => y =
2
x
2
4
=>
dy
dx
=
2x
4
=
x
2
x =4y
(1,2)
(h,k)
2
Slope of the normal is the negative reciprocal of the slope of the tangent. Thus, the slope of normal is
-2
x =4y
x
.
2
(h,k)
(1,2)
Consider a point of intersection of the curve and normal as (h,k). Thus the slope of normal at (h,k) is
-2
x =4y
h
.
2
(h,k)
(1,2)
Also, equation of normal passing through (h, k) is (y-y ) = m (x-x )
x =4y
1
1
=> (y-k) =
-2
h
( x-h)
2
(h,k)
(1,2)
Further, the normal passes through (1, 2) and satisfies the above equation.
x =4y
=> (2-k) =
-2
h
(1-h)
=> k = 2 +
2
h
(1-h)
2
(h,k)
(1,2)
Since, (h,k) also lies on the given curve, we can write the equation of curve as k = .
x =4y
2
h
4
2
(h,k)
(1,2)
Substitute this in the previous equation:
x =4y
=>
2
h
2
h
4
= 2 +
2
h
(1-h)
2
h
-2
2
h
4
=
=>
3
h
=8 or h = 2 .
∵ k =
2
h
4
=
2
2
4
= 1
2
(h,k)
(1,2)
= 2 +
Substitute h = 2 and k = 1 in the equation of normal (y-k) =
x =4y
-2
h
=>(y-1) =
2
-2
(x-h)
we get
(x-2)
=> y - 1 = -x + 2
=> x+y = 3
,
2
(h,k)
(1,2)