By Team Aakash Byju's | 25th December 2022
3
A.
B.
C.
D.
( 1 , )
1
3
( 2 , )
2
3
( 3 , )
1
3
( 1 , )
1
3
( 1 , )
1
3
Find the slope of tangent and the slope of normal to the given curve 3y = 6x − 5x .
3
3y = 6x − 5x
3
dy
dx
=
2 - 5x
2
(h,k)
Slope of tangent = 2 − 5x and
Slope of normal =
1
5x -2
2
(∵ slope of normal = -( )
2
1
slope of tangent
(h,k)
Let P(x , y ) be the point that we are asked to find. Then the equation of the normal line at this point is,
1
1
1
5x -2
2
y − y =
1
1
(x − x )
(h,k)
1
Also, normal passes through the origin i.e., (0, 0) (given in the question)
1
5x -2
2
=> 0 − y =
1
1
(0 − x )
x
5x -2
2
=> y =
1
1
1
(h,k)
1
We got the equation of normal passing through the origin and a point (x , y ). Use all points given in the option to check which option is correct.
1
1
i.e., substitute each point in the equation y =
1
x
5x -2
2
1
1
3y = 6x − 5x
3
(h,k)
Normal
A:
( 1 , )
1
3
=> y =
x
5x -2
2
1
1
1
=> =
1
5(1) -2
2
1
3
=> =
1
3
1
3
,
which is true.
3y = 6x − 5x
3
(h,k)
Normal
B:
( 2 , )
2
3
=> y =
x
5x -2
2
1
1
1
=> =
2
5(2) -2
2
2
3
=> =
2
3
2
18
which is not true.
=> =
2
3
2
20 -2
,
=> =
2
3
1
9
3y = 6x − 5x
3
(h,k)
Normal
C:
( 3 , )
1
3
=> y =
x
5x -2
2
1
1
1
=> =
3
5(3) -2
2
1
3
=> =
1
3
3
45-2
which is not true.
=> =
1
3
3
5(9) -2
,
=> =
1
3
3
43
3y = 6x − 5x
3
(h,k)
Normal
D:
( 1 , )
1
2
=> y =
x
5x -2
2
1
1
1
=> =
1
5(1) -2
2
1
2
=> =
1
2
1
3
which is not true.
=> =
1
2
1
5 -2
Thus, option A
( 1 , )
1
3
is the correct answer.
3y = 6x − 5x
3
(h,k)
Normal