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Joint Entrance Examination
Mathematics
Application of Derivatives
Question 82
Question 82 of 212
Medium
MCQ
The normal to the curve \[x = a\left( {\cos \theta + \theta \sin \theta } \right),y = a\left( {\sin \theta - \theta \cos \theta } \right)\] at any point \[\theta\, '\] is such that
A
it passes through the origin
B
it makes an angle \[{\pi \over 2} + \theta \] with the \[x\]-axis
C
it passes through \[\left( {a{\pi \over 2}, - a} \right)\]
D
it is at a constant distance from the origin
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Official solution
\[x = a\left( {\cos \theta + \theta \sin \theta } \right)\] \[ \Rightarrow {{dx} \over {d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right)\] \[ \Rightarrow {{dx} \over {d\theta }} = a\theta \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\] \[y = a\left( {\sin \theta - \theta \cos \theta } \right)\] \[{{dy} \over {d\theta }} = a\left[ {\cos \theta - \cos \theta + \theta \sin \theta } \right]\] \[ \Rightarrow {{dy} \over {d\theta }} = a\theta \sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)\] From equations
(
1
)
(1)
(
1
)
and
(
2
)
,
(2),
(
2
)
,
we get \[{{dy} \over {dx}} = \tan \theta \Rightarrow \] Slope of normal \[ = - \cot \,\theta \] Equation of normal at \['\theta '\] is \[y - a\left( {\sin \theta - \theta \cos \theta } \right)\] \[ = - \cot \theta \left( {x - a\left( {\cos \theta + \theta \sin \theta } \right)} \right.\] \[ \Rightarrow y\sin \theta - a{\sin ^2}\theta + a\,\,\theta \cos \theta \sin \theta \] \[ = - x\cos \theta + a{\cos ^2}\theta + a\,\theta \sin \theta \cos \theta \] \[ \Rightarrow x\cos \theta + y\sin \theta = a\] Clearly this is an equation of straight line - which is at a constant distance
′
a
′
'a'
′
a
′
from origin.
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