JEE Advanced · Mathematics · 17. Properties of Triangles
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
- A
\(b+c=4 a\)
- B
\(b+c=2 a\)
- C
locus of point \(A\) is an ellipse
- D
locus of point \(A\) is a pair of straight line
Answer & Solution
Correct Answer
(C)
locus of point \(A\) is an ellipse
Step-by-step Solution
Detailed explanation
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.
\[
\begin{aligned}
& \Rightarrow 2 \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)=4 \sin ^2 \frac{A}{2} \\
& \Rightarrow 2 \sin \frac{A}{2}\left[\cos \left(\frac{B-C}{2}\right)-2 \sin \frac{A}{2}\right]=0 \\
& \Rightarrow \cos \left(\frac{B-C}{2}\right)-2 \cos \left(\frac{B+C}{2}\right)=0 \\
& \Rightarrow-\cos \frac{B}{2} \cos \frac{C}{2}+3 \sin \frac{B}{2} \sin \frac{C}{2}=0 \\
& \Rightarrow \quad \tan \frac{B}{2} \tan \frac{C}{2}=\frac{1}{3} \\
& \Rightarrow \quad \sqrt{\frac{(s-a)(s-c)}{s(s-b)} \cdot \frac{(s-b)(s-a)}{s(s-c)}}=\frac{1}{3} \\
& \Rightarrow \quad \frac{s-a}{s}=\frac{1}{3} \Rightarrow 2 s=3 a \\
& \Rightarrow \quad b+c=2 a
\end{aligned}
\]

\(\therefore\) Locus of \(A\) is an ellipse.
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