In a $\triangle A B C$ with fixed base $B C$, the vertex $A$ moves such that $\cos B+\cos C=4 \sin ^2…

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
  1. $b+c=4 a$
  2. $b+c=2 a$
  3. locus of point $A$ is an ellipse
  4. locus of point $A$ is a pair of straight line

Solution

Given, $\cos B+\cos C=4 \sin ^2 \frac{A}{2}$ $ \begin{aligned} & \Rightarrow 2 \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)=4 \sin ^2 \frac{A}{2} \\ & \Rightarrow 2 \sin \frac{A}{2}\left[\cos \left(\frac{B-C}{2}\right)-2 \sin \frac{A}{2}\right]=0 \\ & \Rightarrow \cos \left(\frac{B-C}{2}\right)-2 \cos \left(\frac{B+C}{2}\right)=0 \\ & \Rightarrow-\cos \frac{B}{2} \cos \frac{C}{2}+3 \sin \frac{B}{2} \sin \frac{C}{2}=0 \\ & \Rightarrow \quad \tan \frac{B}{2} \tan \frac{C}{2}=\frac{1}{3} \\ & \Rightarrow \quad \sqrt{\frac{(s-a)(s-c)}{s(s-b)} \cdot \frac{(s-b)(s-a)}{s(s-c)}}=\frac{1}{3} \\ & \Rightarrow \quad \frac{s-a}{s}=\frac{1}{3} \Rightarrow 2 s=3 a \\ & \Rightarrow \quad b+c=2 a \end{aligned} $
$\therefore$ Locus of $A$ is an ellipse

Asked in: JEE Advanced 2009 (Paper 1)

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