The coordinates of the vertices `Ba n dC` of a triangle `A B C` are (2, 0) and (8, 0), respectively. Vertex `A` is moving in such a way that `4tanB/2tanC/2=1.` Then find the locus of `A`

Options

A. `((x-5)^(2))/(25)+(y^(2))/(16)=1`
B. `((x-5)^(2))/(16)+(y^(2))/(25) =1`
C. `((x-5)^(2))/(25)+(y^(2))/(9) =1`
D. `((x-5)^(2))/(9)+(y^(2))/(25)=1`

Answer

A

Explanation

`4 tan.(B)/(2) tan. (C )/(2) =4 sqrt(((s-a)(s-c))/(s(s-b))) sqrt(((s-a)(s-b))/(s(s-c))) =1`
`rArr 4 ((s-a))/(s) =1`
`rArr s = (4a)/(3) = 4 xx (6)/(3) =8`
Now `2s =a +b+c = 16`
`rArr b +c = 10`
Hence locus is an ellipse having center `-= (5,0)`
`2ae = 6`
And `2a = 10`
`b^(2) = a^(2) - a^(2)e^(2) =25 -9 = 16`
`:.` Equation of ellipse is `((x-5)^(2))/(25) +(y^(2))/(16) =1`,