IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Werewolf

This is your question 3
Let the parametric equation of the curve be x=a(cos heta

+

sin

)....(1) and y=a(sin heta

-

cos

)......(2)

Multiplying equation 1 by cos heta

and equation 2 by sin heta

and adding the two we get,
xcos heta

+ysin

=a.....(3)

Slope of line 3 = -cot heta

, this means slope of normal to this line is tan heta

Thus eqn. of the required normal is (y -a(sin heta

-

cos

) )=tan

(x-a(cos

+

sin

))
=> eqn of normal on simplification is a heta

=sin

x-cos

y.....(4)

Equation 4 satisfies the required conditions in the question.