The point to which the origin should be shift | vedclass

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(D) Given equation is $y^2-6y-4x+13=0$. Let the origin be shifted to $(h, k)$. Then $y$ becomes $y+k$ and $x$ becomes $x+h$. Substituting these in the

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$1, 3$

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(D) Given equation is $y^2-6y-4x+13=0$. Let the origin be shifted to $(h, k)$. Then $y$ becomes $y+k$ and $x$ becomes $x+h$. Substituting these in the equation: $(y+k)^2 - 6(y+k) - 4(x+h) + 13 = 0$. Expanding this: $y^2 + 2ky + k^2 - 6y - 6k - 4x - 4h + 13 = 0$. Grouping terms: $y^2 + (2k-6)y - 4x + (k^2 - 6k - 4h + 13) = 0$. For the equation to have no term in $y$,the coefficient of $y$ must be zero: $2k - 6 = 0 \implies k = 3$. For the equation to have no constant term,the constant part must be zero: $k^2 - 6k - 4h + 13 = 0$. Substituting $k = 3$: $(3)^2 - 6(3) - 4h + 13 = 0$. $9 - 18 - 4h + 13 = 0$. $-9 - 4h + 13 = 0$. $4 - 4h = 0 \implies h = 1$. Thus,the origin should be shifted to $(1, 3)$.

The point to which the origin should be shifted so that the equation $y^2-6y-4x+13=0$ will not contain any term in $y$ and the constant term,is

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