Question

Coordinates of the vertices $B$ and $C$ of a triangle ABC are $(2,0)$ and $(8,0)$ respectively. The vertex A is varying in such a way that $4$ $\tan \frac{B}{2}$ and $\frac{C}{2}=1.$ Then locus of $A$ is

• A. $\frac{(x-5{)}^2}{25}+\frac{y^2}{16}=1$
• B. $\frac{(x-5{)}^2}{16}+\frac{y^2}{25}=1$
• C. $\frac{(x-5{)}^2}{25}+\frac{y^2}{9}=1$
• D. $\frac{(x-5{)}^2}{9}+\frac{y^2}{25}=1$

Answer 1

Answer: (A)

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

Given that A,B and C are vertices of a $△$ABC

The coordinate of B and C are (2,0) and (8,0)

$\therefore$ $a=8-\alpha =6$ units

we can assume the coordinate of A to be $(x,y)$

$\Rightarrow b=\sqrt{(x-2{)}^2+y^2},c=\sqrt{(x-8{)}^2+y^2}$

From half angle formulas

$\tan \frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\text{and}\tan \frac{c}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$

$s=\frac{a+b+c}{2}$ i.e, perimeter.

Given that $4\tan \frac{B}{2}\tan \frac{C}{2}=1$

$\Rightarrow 4\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=1$

Squaring on both side,

$16\left[\frac{(s-a)(s-c)}{s(s-b)}\right]\left[\frac{(s-a)(s-b)}{s(s-c)}\right]=1=\frac{16(s-b{)}^2}{s^2}\left[\frac{(s-c)(s-a)}{(s-a)(s-c)}\right]=1\Rightarrow 16(s-b{)}^2=s^2$

taking root,$4(s-b)=s$

$\Rightarrow$ 3s-4b=0

Substituting value of$s=s=\frac{a+b+c}{2}$

$3\left[\frac{a+b+c}{2}\right]-4b=0\Rightarrow 3a-5b+3c=0$

substituting the values of a,b and c

$3\times 6-5\sqrt{(x-8{)}^2+y^2}+3\sqrt{(x-2{)}^2+y^2}=0\therefore 18-5\sqrt{(x-8{)}^2+y^2}+3\sqrt{(x-2{)}^2+y^2}=0$,

The Locus of A