(a, b) is the point to which the origin has t | vedclass

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(B) Step $1$: Shifting the origin to $(h, k)$ to remove first-degree terms from $2x^2 - 3xy + 4y^2 + 5y - 6 = 0$. Let $x = X + h$ and $y = Y + k$. Sub

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$60$

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(B) Step $1$: Shifting the origin to $(h, k)$ to remove first-degree terms from $2x^2 - 3xy + 4y^2 + 5y - 6 = 0$. Let $x = X + h$ and $y = Y + k$. Substituting these into the equation,the coefficients of $X$ and $Y$ must be zero. $4h - 3k = 0$ and $-3h + 8k + 5 = 0$. Solving these,we get $h = -\frac{15}{23}$ and $k = -\frac{20}{23}$. The origin is shifted to $(h, k)$,so the point $(a, b)$ is $(-\frac{15}{23}, -\frac{20}{23})$. Step $2$: Substitute $a = -\frac{15}{23}$ and $b = -\frac{20}{23}$ into the equation $ax^2 + 23abxy + by^2 = 0$. $-\frac{15}{23}x^2 + 23(-\frac{15}{23})(-\frac{20}{23})xy - \frac{20}{23}y^2 = 0$. Multiplying by $-23$,we get $15x^2 - 300xy + 20y^2 = 0$,or $3x^2 - 60xy + 4y^2 = 0$. Step $3$: To remove the $xy$-term by rotating axes by angle $	heta$,the formula is $	an 2	heta = \frac{B}{A - C}$,where the equation is $Ax^2 + Bxy + Cy^2 = 0$. Here $A = 3, B = -60, C = 4$. $	an 2	heta = \frac{-60}{3 - 4} = \frac{-60}{-1} = 60$.

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