In a triangle ABC with fixed base BC, the vertex A moves such that `cos B+cosC=4sin^(2)(A/2)`.
If a,b,c denote the lengths of the triangle opposite to the angles A, B and C respectively, then

Options

A. b+c=4a
B. b+c=2a
C. locus of point A is an ellipse
D. locus of point A is a pair of straight lines

Answer

C

Explanation

We have,
`cosB+cosC=4sin^(2)A/2`
`rArr2cos((B+C)/2)cos((B-C)/2)=4sin^(2)A/2`
`rArrcos((B-C)/2)=2sinA/2`
`rArr2cosA/2cos((B-C)/2)=4sinA/2cosA/2`
`rArr2sin((B+C)/2)cos((B-C)/2)=2sinA`
`rArrsin B + sin C = 2 sin A`
`rArrb+c=2a` `[because(sinA)/a=(sinB)/b=(sinC)/c]`
So, option (b) is correct.
Again, b + c = 2a
`rArrAC+AB=2BC`
`rArr`Point A moves in such a way that the sum of its distances from points B and C is constant and is equal to 2BC.
`rArr`Locus of point A is an ellipse having its foci at B and C.
So, option (c) is correct.