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Question : 67 of 160
Marks:
+1 ,
-0
Solution:
Given,
x = a ( cos ‌ θ + θ ‌ sin ‌ θ ) and
y = a ( sin ‌ θ − θ ‌ cos ‌ θ ) Now,
‌ = ‌ ⋅ ‌ = tan ‌ θ = Slope of tangent
∴ ‌ ‌ ‌ = ‌ d ( a ‌ cos ‌ θ + a θ ‌ sin ‌ θ ) d θ
‌ ‌ = ‌ + ‌ ‌ ‌ = a ( − sin ‌ θ ) + a ‌ Now,
‌ = ‌ d ( a ‌ sin ‌ θ − a θ ‌ cos ‌ θ ) d θ
= ‌ − ‌ = a ‌ cos ‌ θ − a ‌ Now,
‌ = ‌ a θ ‌ sin ‌ θ a θ ‌ cos ‌ θ
= tan ‌ θ ∴ Required equation of Normal
⇒ sin ‌ θ ( y − a ‌ sin ‌ θ + a θ ‌ cos ‌ θ ) = − cos ‌ θ ( x − a ‌ cos ‌ θ − a θ ‌ sin ‌ θ ) ⇒ ‌ ‌ y ‌ sin ‌ θ − a sin 2 θ + x ‌ cos ‌ θ − a cos 2 θ
= a θ ‌ sin ‌ θ ‌ cos ‌ θ − a θ ‌ sin ‌ θ ‌ cos ‌ θ ⇒ ‌ ‌ y ‌ sin ‌ θ + x ‌ cos ‌ θ − a ( 1 ) = 0 ⇒ ‌ ‌ x ‌ cos ‌ θ + y ‌ sin ‌ θ − a = 0 So, perpendicular distance from origin,
d ‌ ‌ = ‌ | cos ‌ θ ( 0 ) + sin ‌ θ ⋅ ( 0 ) − a | √ cos 2 θ + sin 2 θ
‌ ‌ = ‌ = ‌ ⇒ ‌ ‌ d ‌ ‌ = a © examsiri.com
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