MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11
        
I have seen a previous solution which is not very clear for me. So can someone send me a clearer answer to the attached question 53?
one year ago

Answers : (1)

Arun
23340 Points
							
Dear student
 
Equation tangent to H at P is xx1-yy1=1
l=(x1+x2+1/x1)/3 and m=y1/3=sqrt(x1^2-1)/3
 
now dy/dx(H) at P =dy/dx(S) at P
=>x1/y1 =(x2-x1)/y1
Therefore, l=x1+1/3x1
so, dl/dx1=1- 1/(3x1^2) , 
dm/dx1=1/3
and dm/dx1 =(1/3)*x1/(sqrt(x1^2-3)
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 53 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

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