IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Werewolf

This is your last question

Let the parametric equation of the curve be x=acos⁶ heta

and y=asin⁶ heta

dy/dx=slope of tangent = -(tan heta

)⁴

Now the equation of the tangent is (y-asin⁶ heta

)=-(tan

)⁴(x-acos⁶ heta

)
Put x=0 get y intercept= asin⁴ heta

Put y=0 and get x intercept =acos⁴ heta

Now the point given of tangency is (a/8,a/8) , using this we get a/8=acos⁶ heta

and a/8=asin⁶ heta

=>

=45^o=>squares of intercepts are a²/16 and a²/16

Now according to given condition we have a²/16 + a²/16 =2
=> a=4