MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
Mayank Bhardwaj Grade: 12
        

The prependicular from the origin to the tangent at any pt. on curve is equal to abscissa of that pt Find curve ? 

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear Mayank


let the point on the curve is (x1,y1)


slope of the tangent at this point  is (dy/dx)1


equation of tangent


y -y1 =(dy/dx)1 (x-x1)


y -x(dy/dx)+x1(dy/dx)-y1 =0


length of normal from origin


|x1(dy/dx)-y1| /√{(dy/dx)1+ 1}  = |x1|


|x1(dy/dx)-y1|   = |x1|√{(dy/dx)1+ 1}


square and simplyfy


  {y12 -x12}/2x1y1  =(dydx)1


and for general point


 {y2 -x2}/2xy  =(dydx)


this is ordinary diferential equation you cab easly solve







 Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed  solution very  quickly.

 We are all IITians and here to help you in your IIT JEE preparation.

All the best.
 
Regards,
Askiitians Experts
Badiuddin

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details