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Q.

If p and q are lengths of perpendiculars from the origin on the tangent and the normal to the curve x2/3+y2/3=a2/3, then  4p2+q2=

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a

2a2

b

a

c

5 a2

d

a2

answer is C.

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Detailed Solution

x=acos3θ,y=asin3θ dydx=sinθcosθ
Equation of tangent 
yasin3θ=sinθcosθ(xacos3θ) xsinθ+ycosθ=acosθsinθ
P=length of el from (0,0) is 
|c|a2+b2=|asinθcosθ|sin2θ+cos2θ P=12a.sin2θ
Equation of normal 
yasin3θ=cosθsinθ(xacos3θ) xcosθysinθ=acos2θ
Q=length of elfrom (0,0) is 
|acos2θ|cos2θ+sin2θ q=acos2θ
4p2+q2=4.14.a2sin22θ+a2cos22θ=a2(1)=a2

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