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State whether the statements are true or false.

Equation of the line passing through the point (a cos3θ, a sin3θ) and perpendicular to the line x sec θ + y cosec θ = a is x cos θ – y sin θ = a sin 2θ.

Answers (1)

Let the equation of line y=mx+c….(i) 

 So, slope of the above equation is m  

Given equation of line is x secθ+y cosecθ=a sin 2θ.

\frac{\sec \theta x }{cosec \theta }+y=-\frac{a \sin 2 \theta}{cosec \theta }

 Since the slope of the equation is m'=y=-\frac{\sec \theta }{cosec \theta }

  Given that equation (i) is perpendicular to x secθ+y cosec θ=a sin 2θ

 m*m'= -1

  m*\left (-\frac{\sec \theta }{cosec \theta } \right )=-1

m=\frac{cosec \theta }{\sec \theta }

   Putting the value of m in equation (i)we get y=\frac{cosec \theta }{\sec \theta }x+c

y=\frac{cosec \theta +c\left ( sec \theta \right ) }{\sec \theta }

 ysecθ=x cosec θ+c secθ  

x cosec θ-y secθ=k….(ii) 

 If equation (ii) passes through the point (a cos3θ, a sin3θ) 

  (a cos3θ )cosecθ-(a sin3θ) secθ=k 

\frac{a\cos ^{3} \theta}{\sin \theta}- \frac{a\sin ^{3} \theta}{\cos \theta}=x \, cosec\, \, \theta -y\sec \theta

\frac{a\cos ^{4} \theta -a\sin ^{4}\theta}{\sin \theta \cos \theta}=\frac{x}{\sin \theta }-\frac{x}{\cos \theta }   

 a[(cos2θ-sin2θ)(cos2θ+sin2θ)]=x cosθ-ysinθ  

 a[cos2θ-sin2θ]=xcosθ-ysinθ 

 a[cos2θ]=xcos θ-y sinθ 

  The given statement is FALSE

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