Question

In $XY$ plane, the coordinates of two points $B$ and $C$ are $(2,0)$ and $(8,0)$ respectively. Another point $A$ is varying in such a way that $4.\tan \frac{B}{2}.\tan \frac{C}{2}=1$. 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. $\frac{(x+5{)}^2}{25}+\frac{y^2}{9}=1$

Answer 1

The correct option is A $\frac{(x-5{)}^2}{25}+\frac{y^2}{16}=1$

In a triangle $ABC$, $tan\frac{B}{2}=\frac{\Delta }{s(s-c)}$.
Given that $4\tan \frac{B}{2}\tan \frac{C}{2}=1$
$\Rightarrow 4\frac{\Delta }{(s-b)s}\times \frac{\Delta }{(s-c)s}=1$
$\Rightarrow \frac{4\times {\Delta }^2(s-a)}{(s-a)(s-b)(s-c)s\times s}=1$
$\Rightarrow \frac{s-a}{s}=\frac{1}{4}$
$\Rightarrow 4b+4c-4a=a+b+c$
$\Rightarrow 3(b+c)=5a$
Let, point $A$ has coordinates $(h,k)$. Then from above equation we get:
$\sqrt{(h-2{)}^2+(k-0{)}^2}+\sqrt{(h-8{)}^2+k^2}=\frac{5}{3}\times 6$
From the above equation we can notice that , sum of distances of $(h,k)$ from two fixed points $(2,0)$ and $(8,0)$ is a constant. Hence, the locus of $(h,k)$ is an ellipse.
sum of distances $=10=2\times$ semi-major axis.
$\Rightarrow$ semi-major axis $a^{\prime }=5$
Center of ellipse is midpoint of two fixed points: $(5,0)$
Also, distance between focii is equal to $6=\sqrt{(a^{\prime }{)}^2-(b^{\prime }{)}^2}$, where $b^{\prime }$ is semi-minor axis.
$\Rightarrow b^{\prime }=4$
Equation of ellipse is:
$\frac{(x-5{)}^2}{25}+\frac{y^2}{16}=1$
Hence, option 'A' is correct.