If (h, k) is the new origin to be chosen to e | vedclass

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(D) Given equation: $S \equiv 2x^2 - xy - y^2 - 3x + 3y = 0$. To eliminate first-degree terms,we find the new origin $(h, k)$ by solving $\frac{\parti

Vedclass · Accepted Answer

$-\frac{h}{3k}$

Vedclass · Answer

(D) Given equation: $S \equiv 2x^2 - xy - y^2 - 3x + 3y = 0$. To eliminate first-degree terms,we find the new origin $(h, k)$ by solving $\frac{\partial S}{\partial x} = 0$ and $\frac{\partial S}{\partial y} = 0$. $\frac{\partial S}{\partial x} = 4x - y - 3 = 0$ and $\frac{\partial S}{\partial y} = -x - 2y + 3 = 0$. Solving these,we get $x = 1, y = 1$,so $(h, k) = (1, 1)$. To eliminate the $xy$-term by rotation,we use the formula $	an 2	heta = \frac{B}{A - C}$,where the equation is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$. Here $A = 2, B = -1, C = -1$. Thus,$	an 2	heta = \frac{-1}{2 - (-1)} = \frac{-1}{3}$. Since $h = 1$ and $k = 1$,we have $h = k$,so $	an 2	heta = -\frac{h}{3k} = -\frac{1}{3}$. Therefore,the correct option is $D$.

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