0

Guest

Courses New

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.

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.

Grade:

1 Answers

vikas askiitian expert
509 Points
12 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

Think You Can Provide A Better Answer ?

Other Related Questions on Differential Calculus

View all Questions »

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More