By shifting the origin to the point (-1, 2) t | vedclass

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(D) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y)$.The translation of axes to the origin $(-1, 2)$ implies $x = X - 1$ a

Vedclass · Accepted Answer

$c-5(a+b)$

Vedclass · Answer

(D) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y)$.The translation of axes to the origin $(-1, 2)$ implies $x = X - 1$ and $y = Y + 2$.Substituting these into the equation $2x^2 - xy + y^2 - 3x + 4y - 5 = 0$:$2(X-1)^2 - (X-1)(Y+2) + (Y+2)^2 - 3(X-1) + 4(Y+2) - 5 = 0$$2(X^2 - 2X + 1) - (XY + 2X - Y - 2) + (Y^2 + 4Y + 4) - 3X + 3 + 4Y + 8 - 5 = 0$$2X^2 - 4X + 2 - XY - 2X + Y + 2 + Y^2 + 4Y + 4 - 3X + 4Y + 11 - 5 = 0$$2X^2 - XY + Y^2 + (-4 - 2 - 3)X + (1 + 4 + 4)Y + (2 + 2 + 4 + 11 - 5) = 0$$2X^2 - XY + Y^2 - 9X + 9Y + 14 = 0$Comparing this with $aX^2 + 2hXY + bY^2 + 2gX + 2fY + c = 0$,we get:$a = 2, 2h = -1 \implies h = -0.5, b = 1, 2g = -9 \implies g = -4.5, 2f = 9 \implies f = 4.5, c = 14$.Now,calculate $2(f + g + h) = 2(4.5 - 4.5 - 0.5) = 2(-0.5) = -1$.Checking the options,$c - 5(a + b) = 14 - 5(2 + 1) = 14 - 15 = -1$.Thus,the correct option is $D$.

By shifting the origin to the point $(-1, 2)$ through translation of axes,if $ax^2+2hxy+by^2+2gx+2fy+c=0$ is the transformed equation of $2x^2-xy+y^2-3x+4y-5=0$,then $2(f+g+h)=$

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