When the origin is shifted to the point ((3/2), (3/2)) by the translation of coordinate axes.then the transformed equation of 32 x2+8 x y+32 y2-108 x-108 y+99=0 is

Question

When the origin is shifted to the point ((3/2), (3/2)) by the translation of coordinate axes.then the transformed equation of 32 x2+8 x y+32 y2-108 x-108 y+99=0 is (A) 72 X2+56 Y2-63=0 (B) X2-14 X Y-7 Y2-2=0 (C) 32 X2-16 X Y+32 Y2-225=0 (D) 32 X2+8 X Y+32 Y2-63=0

Tardigrade Admin · Accepted Answer

Substituting x=X+(3/2), y=Y+(3/2) in the equation 32 x2+8 x y+32 y2-108 x-108 y+99=0 we get 32(X+(3/2))2+8(X+(3/2))(Y+(3/2))+32(Y+(3/2))2 -108(X+(3/2))-108(Y+(3/2))+99=0 ⇒ 32[X2+3 X+(9/4)]+8((2 X+3/2))((2 Y+3/2)) +32[Y2+3 Y+(9/4)]-108 X-162-108 Y-162+99=0 ⇒ 32 X2+96 X+72+2(4 X Y+6 X+6 Y+9)+32 Y2 +96 Y+72-108 X-162-108 Y-162+99=0 ⇒ 32 X2+32 Y2+8 X Y-63=0

Q. When the origin is shifted to the point $(\frac{3}{2}, \frac{3}{2})$ by the translation of coordinate axes.then the transformed equation of $32 x^{2} + 8 x y + 32 y^{2} - 108 x - 108 y + 99 = 0$ is

$72 X^{2} + 56 Y^{2} - 63 = 0$

$X^{2} - 14 X Y - 7 Y^{2} - 2 = 0$

$32 X^{2} - 16 X Y + 32 Y^{2} - 225 = 0$

$32 X^{2} + 8 X Y + 32 Y^{2} - 63 = 0$

Q. When the origin is shifted to the point (23​,23​) by the translation of coordinate axes.then the transformed equation of 32x2+8xy+32y2−108x−108y+99=0 is

72X2+56Y2−63=0

X2−14XY−7Y2−2=0

32X2−16XY+32Y2−225=0

32X2+8XY+32Y2−63=0

Q. When the origin is shifted to the point $(\frac{3}{2}, \frac{3}{2})$ by the translation of coordinate axes.then the transformed equation of $32 x^{2} + 8 x y + 32 y^{2} - 108 x - 108 y + 99 = 0$ is

$72 X^{2} + 56 Y^{2} - 63 = 0$

$X^{2} - 14 X Y - 7 Y^{2} - 2 = 0$

$32 X^{2} - 16 X Y + 32 Y^{2} - 225 = 0$

$32 X^{2} + 8 X Y + 32 Y^{2} - 63 = 0$