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If the equation x squared-4x-6y+10=0 is transformed to X squared+AY=0,the axes remaining parallel.Find the new coordinates of the point where origin is shited and the value of A?

$T h e g i v e n e q u a t i o n i s, x^{2} - 4 x - 6 y + 10 = 0 ..... (1) L e t t h e n e w o r i g i n i s (h, k) L e t (x, y) b e t h e c o - o r d i n a t e o f a n y p o int r e f e r r e d t o o r i g i n a l a x e s . A n d, (x', y') b e t h e c o - o r d i n a t e o f t h e s a m e p o int w . r . t . n e w a x e s, t h e n, x = x' + h y = y' + k T h u s, e q u a t i o n (1) b e c o m e s, {(x' + h)}^{2} - 4 (x' + h) - 6 (y' + k) + 10 = 0 \Rightarrow ({x'}^{2} + h^{2} + 2 x' h) - 4 x' - 4 h - 6 y' - 6 k + 10 = 0 \Rightarrow {x'}^{2} + h^{2} + 2 x' h - 4 x' - 4 h - 6 y' - 6 k + 10 = 0 \Rightarrow {x'}^{2} + x' (2 h - 4) - 6 y' + (h^{2} - 4 h - 6 k + 10) = 0 B u t i t i s g i v e n t h a t t h e t r a n s f o r m e d e q u a t i o n i s i n t h e f o r m o f X^{2} + A Y = 0 . B y c o m p a r i n g b o t h e q u a t i o n s, w e h a v e, A = - 6 A n d, 2 h - 4 = 0 \Rightarrow h = 2 A n d, h^{2} - 4 h - 6 k + 10 = 0 \Rightarrow {(2)}^{2} - 4 \times 2 - 6 k + 10 = 0 \Rightarrow 4 - 8 - 6 k + 10 = 0 \Rightarrow k = 1 T h u s, t h e o r i g i n i s s h i f t e d t o (2, 1) a n d A = - 6$

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