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CBSE Class 11-science Answered

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Asked by dra | 24 Feb, 2024, 06:35: AM
answered-by-expert Expert Answer

Let the equation of the curve when origin is at (0, 0) is

a X2 + b X Y + c Y2 + d X  + e Y + f = 0  ..................... (1)

when origin is shifted to (2, 3 ) , then above equation of curve with respect to new origin becomes

a ( X - 2 ) 2 + b (X-2) ( Y - 3 ) + c ( Y-3)2 + d (X - 2 )  + e ( Y - 3 ) + f = 0

a X2 + b X Y + c Y2 + (-4a-3b+d ) X  + (-2b-6c+e ) Y + ( 4a + 6b +9c-2d-3e+f ) = 0 .....................................(2)

Let us compare the above eqn.(2) with the given equation

x2 + 3 x y - 2 y2 + 17 x  -7 y - 11 = 0  ............................................. (3)

By comparing the respective coefficients of eqn.(2) and eqn.(3) , we get

a = 1  ; b = 3 ;  c = -2

-4a -3b+d = 17  

By substituting a and b in above equation, we get d = 30 ;

-2b-5c+e = -7

By substituting b and c in above equation, we get e = -11  ;

( 4a + 6b +9c-2d-3e+f ) = -11

By substituting a, b, c , d and e in above equation, we get f = 12  ;

Hence equation of curve before shifting th origin is

X2 + 3 X Y - 2 Y2 + 30 X  - 11 Y + 12 = 0

Answered by Thiyagarajan K | 24 Feb, 2024, 12:59: PM
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