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A point moves so that the sum o its distances from the point (k,0) and (-k,0) is 2a. prove that its locus is : X^2/A^2 + Y^2/B^2 =1 where B^2 =A^2 -K^2.  
Asked by chalaunesuresh32 | 22 Feb, 2019, 08:08: AM
answered-by-expert Expert Answer
Let coordinate of P be (x,y). We are given that,  PA +PB = 2a
 
begin mathsize 12px style square root of open parentheses x plus k close parentheses squared plus y squared space end root space plus space square root of open parentheses x minus k close parentheses squared plus y squared space end root space equals space 2 a square root of open parentheses x plus k close parentheses squared plus y squared space end root space equals space 2 a space minus space square root of open parentheses x minus k close parentheses squared plus y squared space end root B y space s q u a r i n g space b o t h space s i d e s colon space open parentheses x plus k close parentheses squared plus y squared space equals space 4 a squared plus open parentheses x minus k close parentheses squared plus y squared space minus space 4 a square root of open parentheses x minus k close parentheses squared plus y squared space end root x squared plus k squared plus 2 x k plus y squared space equals space 4 a squared plus x squared plus y squared plus k squared minus 2 x k minus 4 a square root of open parentheses x minus k close parentheses squared plus y squared end root B y space c a n c e l l i n g space o u t space e q u a t e space t e r m s comma space s q u a r e minus r o o t space t e r m space i s space w r i t t e n space a s square root of open parentheses x minus k close parentheses squared plus y squared end root space equals space fraction numerator 1 over denominator 4 a end fraction open parentheses 4 a squared minus 4 x k close parentheses space equals a space minus fraction numerator x space k over denominator a end fraction B y space s q u a r i n g space b o t h space s i d e s open parentheses x minus k close parentheses squared plus y squared space equals space open parentheses a space minus fraction numerator x space k over denominator a end fraction close parentheses squared x squared plus k squared minus 2 x k plus y squared space equals space a squared plus fraction numerator x squared k squared over denominator a squared end fraction minus 2 x k x squared open parentheses 1 minus k squared over a squared close parentheses plus space y squared equals a squared minus k squared space space space o r space space x squared over a squared plus fraction numerator y squared over denominator open parentheses a squared minus k squared close parentheses end fraction space equals space 1 end style
Answered by Thiyagarajan K | 23 Feb, 2019, 03:55: PM

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