Vikas TU
Last Activity: 7 Years ago
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)) [Ans]
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)