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STD 11 - 10.2 parabola,ellipse,hyperbola — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 10.2 parabola,ellipse,hyperbola4 Marks
MCQ
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 lines

  • ✓
    $(B,C)$
  • B
    $(B,D)$
  • C
    $(A,C)$
  • D
    $(A,D)$
✓

Answer

Correct option: A.
$(B,C)$
a
We know that in triangle $ABC , A + B + C =18$ o degrees

$\therefore B + C =180- A$

$\cos B +\cos C =4 \sin ^2 \frac{ A }{2}$

$2 \cos \frac{ B + C }{2} \cos \frac{ B - C }{2}=4 \sin ^2 \frac{ A }{2}$

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

$2 \cos \frac{ A }{2} \cos \frac{ B - C }{2}=4 \cos \frac{ A }{2} \sin \frac{ A }{2}$

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

$\sin B +\sin C =2 \sin A$

$\Rightarrow b + C =2 a$

$\Rightarrow AC + AB =2 BC$

Now point $A$ moves in such a way that the sum of its distance from points $B$ and $C$ is constant and equal to $2$ $BC$

So its locus is ellipse.

So options $B$ and $C$ are correct.

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