If the greatest value of the term independent of x in expansion of (xsinp + x-1cosp)10 is achieved at P=θ... then the locus of point from which pair of tangents be drawn to x2+y2=4 including an angle θ is
A) x2+y2= 4(4+2√2) B) x2+y2=4(4+√2)
C) 2x2+y2=4(3+2√2) D) x2+y2=4(1+2√2)
1term independent of x = 10C5 sin5p cos5p = 10C5/32 sin52p
max value of sin52p = 1 at θ = π/4
i got the following result .. by drawing a diagram ..
(x - 0)2 + (y - 0)2 = 4(cosec 22.5°)2
x2 + y2 = 4(cosec 22.5°)2
cos 45° =- 2sin2 22.5° + 1
- 1/√2 + 1 = 2sin2 22.5°
sin2 22.5° =( 1 - 1/√2 ) /2
cosec2 22.5° =2/( 1 - 1/√2 )
=2√2/( √2 - 1 ) = 2√2( √2 + 1 ) = 4 + 2√2
x2 + y2 = 4(4 + 2√2 )
A ...
11it is clear that 5th term will be free of x
so
10C5sin5pcos5p the greatest value of this can be found out differentiating it with respect to p
so it would be max at p=pi/4
so max value is 10C5 X 1 /25 =63/8
for pair of tangents of eq of 2 lines is ax2+by2+2hxy+2gx +2fy +c
tanθ=√h2-ab/a+b
put θ=pi/4 to get the requisite answer [1]
1now i got a doubt...
is there any formula to find d locus of d point whose tangents to a given circle S=0 include an angle θ or should it be solved diagramatically...!!???
62It can be done either way :)
not necessarily diagramatically
as mak points out it will not be very easy in this case unless you know
tan (45/2)!
1ur answer is correct ankit... :)
so bhaiya... v dun have any standard formula for dat kya...!!??? [12]