Question

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

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

Answer 1

Answer: (B), (C)

The correct options are
B $b+c=2a$
C locus point A is an ellipse

$\cos B+\cos C=4{\sin }^2\left(\frac{A}{2}\right)$
$\Rightarrow$ $2\cos \frac{B+C}{2}\cos \frac{B-C}{2}=4{\sin }^2\left(\frac{A}{2}\right)$
$\Rightarrow$ $2\cos (\frac{\pi }{2}-\frac{A}{2})\cos \frac{B-C}{2}=4\sin \frac{A}{2}\sin (\frac{\pi }{2}-\frac{B+C}{2})$

$\Rightarrow$ $\frac{\cos \frac{B-C}{2}}{\cos \frac{B+C}{2}}=2$

$\Rightarrow$ $\frac{\cos \frac{B}{2}\cos \frac{C}{2}+\sin \frac{B}{2}\sin \frac{C}{2}}{\cos \frac{B}{2}\cos \frac{C}{2}-\sin \frac{B}{2}\sin \frac{C}{2}}=2$

$\Rightarrow$ $\tan \frac{B}{2}\tan \frac{C}{2}=\frac{1}{3}\left(applyingcomponendoanddividendo\right)$

$\Rightarrow$ $\frac{{\Delta }^2}{s(s-b)s(s-c)}=\frac{1}{3}(\therefore \tan \frac{B}{2}=\frac{\Delta }{s(s-a)})$

$\Rightarrow$ $3s(s-a)(s-b)(s-c)=s^2(s-b)(s-c)$
$\Rightarrow$ $3(s-a)=s\Rightarrow 2s=3a$
$\Rightarrow$ $a+b+c=3a$ $\Rightarrow b+c=2a$

Therefore, sum of distances of a point, A, from two fixed points B and C is a constant. Thus, the locus of A is an ellipse.