Mathematics

Problem

Coordinates of the vertices B and C of a

\Delta ABC

are (2, 0) and (8, 0) respectively. The vertex A is varying in such a way that

4 \tan\left(\frac{B}{2}\right) \cdot \tan\left(\frac{C}{2}\right) = 1

. Then, the locus of A is a.

\frac{(x-5)^2}{25} + \frac{y^2}{16} = 1

\frac{(x-5)^2}{16} + \frac{y^2}{9} = 1

\frac{(x-5)^2}{25} + \frac{y^2}{9} = 1

\frac{(x-5)^2}{16} + \frac{y^2}{25} = 1