In a triangle $A B C$ with fixed base $B C$, the vertex $A$ moves such that $$ \cos B+\cos C=4 \sin ^{2} \frac{A}{2} $$ If $a, b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A, B$ and $C$, respectively, then

(B) $b+c=2 a$; (C) locus of point $A$ is an ellipse

(D) locus of point $A$ is a pair of straight lines

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Properties of Triangle

In a triangle

A B C

with fixed base

B C

, the vertex

A

moves such that

cos B + cos C = 4 sin^{2} \frac{A}{2}

a, b

and

c

denote the lengths of the sides of the triangle opposite to the angles

A, B

and

C

, respectively, then

Answer: (B), (C)

Using the identity

cos B + cos C = 2 cos (\frac{B + C}{2}) cos (\frac{B - C}{2})

and

sin^{2} \frac{A}{2} = \frac{1 - c o s A}{2} = \frac{1 - c o s ( π - B - C )}{2} = \frac{1 + c o s ( B + C )}{2}

:

Also,

4 sin^{2} \frac{A}{2} = 2 (1 - cos A) = 2 (1 + cos (B + C)) = 2 \cdot 2 cos^{2} \frac{B + C}{2} = 4 sin^{2} \frac{A}{2}

From the given condition:

2 cos (\frac{B + C}{2}) cos (\frac{B - C}{2}) = 4 sin^{2} \frac{A}{2} = 2 sin \frac{A}{2} \cdot 2 sin \frac{A}{2}

Since

cos \frac{B + C}{2} = sin \frac{A}{2}

cos (\frac{B - C}{2}) = 2 sin \frac{A}{2}

Using sine rule:

\frac{s i n B + s i n C}{s i n A} = 2

, which gives

b + c = 2 a

(constant).

Since

b + c = 2 a

is constant and

B C

is fixed, the locus of

A

is an ellipse with foci at

B

and

C

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · One or more correct options