The point to which the origin should be shifted in order to remove the x and y terms in the equation
14x2 4xy + 11y2 – 36x + 48y + 41 = 0 is (h,k) then the value of h-k is
Answers
the transformed equation is
14(x′ +h) 2 −4(x ′ +h)(y ′ +k)+11(y ′ +k) 2 −36(x ′ +h)+48(y ′ +k)+41=0
14x′ 2 +28x′ h+14h 2 −4x′ y′ −4x′ k−4y′ h−4hk+11y′ 2 +11k 2 +22y′ k−36x′ −36h+48y ′ +48k+41=0
14x′ 2 +11y′ 2 +x′ (28h−4k−36)+y′ (−4h+22k+48) +(14h 2 −4hk−36h+48k+41)−4x′ y′ =0
In order to remove the first degree term, we must have
28h−4k−36=0 and −4h+22k+48=0
∴h=1 and k=−2
Therefore the answer is (1,-2).
the transformed equation is
14(x′ +h) 2 −4(x ′ +h)(y ′ +k)+11(y ′ +k) 2 −36(x ′ +h)+48(y ′ +k)+41=0
14x′ 2 +28x′ h+14h 2 −4x′ y′ −4x′ k−4y′ h−4hk+11y′ 2 +11k 2 +22y′ k−36x′ −36h+48y ′ +48k+41=0
14x′ 2 +11y′ 2 +x′ (28h−4k−36)+y′ (−4h+22k+48) +(14h 2 −4hk−36h+48k+41)−4x′ y′ =0
In order to remove the first degree term, we must have
28h−4k−36=0 and −4h+22k+48=0
∴h=1 and k=−2
Hence , the answer is (1,-2).