If a normal is drawn at a variable point P of the ellipse (X^2/a^2)+(y^2/b^2)=1,find the maximum distance of the normal from the centre of the ellipse.

general point on ellipse is (x,y)= (acos@,bsin@)

  x=acos@                              &                    y =bsin@

   dx/d@=-asin@                     &                    dy/d@=bcos@

     dy/dx=-bcot@/a

slope of tangent=dy/dx=-(bcot@)/a

slope of normal=-1/dy/dx =atan@/b

now equation of normal at (acos@,bsin@) is

yb-axtan@+a²sin@-b_{²sin@ =0}

distance of this line from center of ellipse(0,0) is

       d=  (a²-b²)sin@/(1+tan²@)^1/2

         =(a²-b²)sin@/sec@=(a²-b²)sin2@ /2

maximum value of sin2@ is 1

so maximum distance is d_max= (a²-b²)/2

Approved