Pratik Tibrewal
Last Activity: 10 Years ago
Let the equation of tangent in slope form be y= mx + a/m
Let point P be (h,k)
hence this point will satisfy the line
there fore: k = mh + a/m or mh^2 - km + a = 0 (a quadratic in 'm' which has two roots m1 (tan(a)and m2 tan(b))
gicen cos(a) . cos(b) = K;
from quadratic product of roots: tan(a) . tan(b) = a/h^2 -------- (i)
sum of roots : tan(a) + tan(b) = k/h^2
now sec^2(a). sec^2(b) =(1+tan^2(a))(1+tan^2(b)) = 1 + tan^2(a) + tan^2(b) + tan^2(a) . tan^2(b) =
= 1 + (tan(a) + tan(b))^2 - 2tan(a) . tan(b) + tan^2(a). tan^2(b)
= 1 + k^2/h^4 - 2a/h^2 + a^2/h^4
now cos (a) . cos(b) = 1/sec(a). sec(b) = K
Thanks and Regards,
Pratik Tibrewal
askiitian faculty,
BTech, IITG
