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Base BC of triangle ABC is fixed and opp...

Base BC of triangle ABC is fixed and opposite vertex A moves in such a way that `tan.(B)/(2)tan.(C)/(2)` is constant. Prove that locus of vertex A is ellipse.

Text Solution

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Given the at in triangle ABC, 'a' is constnat

and `tan.(B)/(2)tan.(C)/(2)= lambda` (constant)
`rArrsqrt(((s-c)(s-c))/(s(s-b)))sqrt(((s-a)(s-b))/(s(s-c)))=lambda`
`rArr (s-a)/(s)=lambda`
`rArr (2s-a)/(a)=a(lambda+1)/(1-lambda)` (constant)
Thus, sum of distance of variable point A from two given fixed ponts Ba dn C is contant , therefore, locus of A is ellipse
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