Application of DerivativeHard
Question
If equation of normal at a point (m2, -m3) on the curve x3 - y2 = 0 is y = 3mx - 4m3, then m2 equals-
Options
A.0
B.1
C.-2/9
D.2/9
Solution
x3 - y2 = 0
3x2 - 2y
= 0
3x2 = 2y
= -
× m
(y + m3) =2/3m (x - m2)
3my + 3m4 = 2x - 2m2
3my = 2x - 3m4 - 2m2
y =
- m3 -
m
but given
3m =
3x2 - 2y
= 03x2 = 2y
= -
× m(y + m3) =2/3m (x - m2)
3my + 3m4 = 2x - 2m2
3my = 2x - 3m4 - 2m2
y =
- m3 -mbut given
3m =
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