Applications-Of-Derivatives Question 447
Question: If $ x=t^{2} $ and $ y=2t $ , then equation of the normal at $ t=1 $ is
[RPET 1996]
Options:
A) $ x+y-3=0 $
B) $ x+y-1=0 $
C) $ x+y+1=0 $
D) $ x+y+3=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ x=t^{2} $ and $ y=2t $ At $ t=1,\ x=1 $ and $ y=2 $ Now $ ( \frac{dy}{dx} )=\frac{dy/dt}{dx/dt}=\frac{2}{2t}=\frac{1}{t} $
$ \Rightarrow $ $ {{( \frac{dy}{dx} )}_{t=1}}=1 $
$ \therefore $ Equation of the normal at (1, 2) is $ y-2=-\frac{1}{\frac{dy}{dx}}(x-1) $
Þ $ y-2=-1(x-1) $
Þ $ x+y-3=0 $ .
BETA
