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# 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.

10 years ago

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@-b2sin@ =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 dmax= (a2-b2)/2