Question

In a triangle ABC with fixed base BC, the vertex A moves such that\[cosB{\text{ }} + {\text{ }}cosC = 4{\sin ^2}\dfrac{A}{2}\]. If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C respectively, then
$
  {\text{A}}{\text{. b + c = 4a}} \\
  {\text{B}}{\text{. b + c = 2a}} \\
  {\text{C}}{\text{. locus of point A is an ellipse}} \\
  {\text{D}}{\text{. locus of point A is a straight line}} \\
$

Answer 1

Hint: The relation between angles and sides is given generally by the sine and cosine rule. So, try to convert the expression \[cosB{\text{ }} + {\text{ }}cosC = 4{\sin ^2}\dfrac{A}{2}\] into sine angles and then tally with the sides. For the locus portion, recall the properties of different conic sections.

Complete step-by-step answer:
Given that point A moves such that
\[cosB{\text{ }} + {\text{ }}cosC = 4{\sin ^2}\dfrac{A}{2}\].
We know that $A + B + C = \pi $ (angle sum property).
$
  cosB{\text{ }} + {\text{ }}cosC = 4{\sin ^2}\dfrac{A}{2} \\
  2\cos \left( {\dfrac{{B + C}}{2}} \right)\cos \left( {\dfrac{{B - C}}{2}} \right) = 4{\sin ^2}\dfrac{A}{2} \\
  \cos \left( {\dfrac{{B + C}}{2}} \right)\cos \left( {\dfrac{{B - C}}{2}} \right) = 2{\sin ^2}\dfrac{A}{2} \\
  \cos \left( {\dfrac{{\pi - A}}{2}} \right)\cos \left( {\dfrac{{B - C}}{2}} \right) = 2{\sin ^2}\dfrac{A}{2} \\
  \cos \left( {\dfrac{\pi }{2} - \dfrac{A}{2}} \right)\cos \left( {\dfrac{{B - C}}{2}} \right) = 2{\sin ^2}\dfrac{A}{2} \\
  \sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{{B - C}}{2}} \right) = 2{\sin ^2}\dfrac{A}{2} \\
  \cos \left( {\dfrac{{B - C}}{2}} \right) = 2\sin \dfrac{A}{2} \\
  {\text{multiply both the sides by cos}}\dfrac{A}{2} \\
{\text{cos}}\dfrac{A}{2}\cos \left( {\dfrac{{B - C}}{2}} \right) = 2\sin \dfrac{A}{2}{\text{cos}}\dfrac{A}{2} \\
  {\text{cos}}\dfrac{A}{2}\cos \left( {\dfrac{{B - C}}{2}} \right) = \sin A \\
  {\text{cos}}\left( {\dfrac{{\pi - (B + C)}}{2}} \right)\cos \left( {\dfrac{{B - C}}{2}} \right) = \sin A \\
  {\text{cos}}\left( {\dfrac{\pi }{2} - \dfrac{{(B + C)}}{2}} \right)\cos \left( {\dfrac{{B - C}}{2}} \right) = \sin A \\
  \sin \left( {\dfrac{{(B + C)}}{2}} \right)\cos \left( {\dfrac{{B - C}}{2}} \right) = \sin A \\
  {\text{multiply both the sides by 2}} \\
  {\text{in order to completely convert both the sides in sine}} \\
  2\sin \left( {\dfrac{{(B + C)}}{2}} \right)\cos \left( {\dfrac{{B - C}}{2}} \right) = 2\sin A \\
  {\text{This becomes}} \\
  {\text{sinB + sinC = 2sinA}} \\
$
From the sine law-
$\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k$
We can say,
$
  \sin B = bk \\
  \sin C = ck \\
  \sin A = ak \\
  So, \\
  b + c = 2a \\
 $
The above expression indicates that as point A is moving, the sum of its distance from two fixed points B and C remains constant.
i.e. we can rewrite the above expression as-
AB + AC = 2BC
 Where b, c, a are the respective lengths of the line segments in the expressions. Now this relation corresponds to the property of an ellipse.
So the locus traced by point A is that of an ellipse.
The correct options are B and C.

Note: It is the property of an ellipse that for any point lying on it, the sum of distances from the two fixed foci of the ellipse remains constant. So, here, the two fixed foci are the points B and C and the moving point A lies on the ellipse.
One can also note that the relation \[cosB{\text{ }} + {\text{ }}cosC = 4{\sin ^2}\dfrac{A}{2}\], holds for any triangle formed within the ellipse, with B and C as foci and A is any point on the ellipse.
The expression b + c = 2a, also indicates the standard definition of an ellipse, $P{F_1} + P{F_2} = 2a$ where a is the length of the major axis.