Nicho priyatham
Last Activity: 9 Years ago
- equation of tangent at an eccentric angle k is
(x/a)cosk+(y/b)sink=0
- equation of tangent at an eccentric angle k+900 is
-(x/a)sink+(y/b)cosk=0
- solving the above equations we get point of intersection (l,m)
l=a(cosk-sink) ; m=b(cosk+sink)
- on further solving we get
2cosk=(l/a+m/b) ; 2sink=(m/b-l/a)
- squaring on both sides and adding we get
(I/a)2+(m/b)2 =2
so locus of( l,m) is option A
PLEASE APPROVE
