show that equation of the line passing through (acos^3 theeta,asin^3theeta) and perpendicular to the line x sec theeta + ycosec theeta = a is x cos theeta -y sin theeta = a cos2 theeta

Dear Student ,Varshini
Please find below the solution to the asked query:

Let co-ordinates of point P be acos3θ, asin3θ and slope of required line L be mGiven equation of line is L1:  xsecθ+ycscθ=aNow slope of any line is given by -Coefficient of xCoefficient of ySlope of L1=m1=-secθcscθWe know that sinθ=1cscθcosθ=1secθtanθ=1cotθtanθ=sinθcosθm1=-sinθcosθm1=-tanθSince L1 and L are perpendicular to each otherm=-1m1m=-1-tanθm=cotθEquation of any line passing through point x1,y1 and having slope m is given byy-y1=mx-x1Hence equation of line L will be given byy- asin3θ=cotθx-acos3θy- asin3θ=cosθsinθx-acos3θsinθy- asin3θ=cosθx-acos3θysinθ- asin4θ=xcosθ-acos4θxcosθ-ysinθ=acos4θ- asin4θxcosθ-ysinθ=acos4θ- sin4θxcosθ-ysinθ=acos2θ2- sin2θ2xcosθ-ysinθ=acos2θ-sin2θcos2θ+sin2θ Since a2-b2=a-ba+bWe know that cos2θ+sin2θ=1 and cos2θ-sin2θ=cos2θxcosθ-ysinθ=acos2θ Hence proved

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