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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.tanB2.tanC2=1. Then locus of A is:

A
(x5)225+y216=1
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B
(x5)216+y225=1
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C
(x5)225+y29=1
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D
(x+5)225+y29=1
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Solution

The correct option is A (x5)225+y216=1
In a triangle ABC, tanB2=Δs(sc).
Given that 4tanB2tanC2=1
4Δ(sb)s×Δ(sc)s=1
4×Δ2(sa)(sa)(sb)(sc)s×s=1
sas=14
4b+4c4a=a+b+c
3(b+c)=5a
Let, point A has coordinates (h,k). Then from above equation we get:
(h2)2+(k0)2+(h8)2+k2=53×6
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× semi-major axis.
semi-major axis a=5
Center of ellipse is midpoint of two fixed points: (5,0)
Also, distance between focii is equal to 6=(a)2(b)2, where b is semi-minor axis.
b=4
Equation of ellipse is:
(x5)225+y216=1
Hence, option 'A' is correct.

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