Transforming to parallel axes through a point | vedclass

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Question

(B) Let the origin be shifted to $(p, q)$ such that the new axes are parallel to the original axes. The transformation equations are $x = x' + p$ and

Vedclass · Accepted Answer

$p = 2, q = -3$

Vedclass · Answer

(B) Let the origin be shifted to $(p, q)$ such that the new axes are parallel to the original axes. The transformation equations are $x = x' + p$ and $y = y' + q$.Substituting these into the given equation $2x^2 + 3xy + 4y^2 + x + 18y + 25 = 0$:$2(x' + p)^2 + 3(x' + p)(y' + q) + 4(y' + q)^2 + (x' + p) + 18(y' + q) + 25 = 0$Expanding the terms:$2(x'^2 + 2px' + p^2) + 3(x'y' + qx' + py' + pq) + 4(y'^2 + 2qy' + q^2) + x' + p + 18y' + 18q + 25 = 0$Grouping the terms by $x'$, $y'$, and constants:$2x'^2 + 3x'y' + 4y'^2 + (4p + 3q + 1)x' + (3p + 8q + 18)y' + (2p^2 + 3pq + 4q^2 + p + 18q + 25) = 0$Comparing this with the given transformed equation $2x^2 + 3xy + 4y^2 = 1$ (or $2x'^2 + 3x'y' + 4y'^2 - 1 = 0$), the coefficients of $x'$ and $y'$ must be zero:$4p + 3q + 1 = 0$ ... $(i)$$3p + 8q + 18 = 0$ ... (ii)Solving these linear equations:From $(i)$, $4p = -3q - 1 \implies p = \frac{-3q - 1}{4}$.Substituting into (ii): $3(\frac{-3q - 1}{4}) + 8q + 18 = 0$$-9q - 3 + 32q + 72 = 0 \implies 23q + 69 = 0 \implies q = -3$.Substituting $q = -3$ into $(i)$: $4p + 3(-3) + 1 = 0 \implies 4p - 8 = 0 \implies p = 2$.Thus, $p = 2, q = -3$.

Transforming to parallel axes through a point $(p, q)$, the equation $2x^2 + 3xy + 4y^2 + x + 18y + 25 = 0$ becomes $2x^2 + 3xy + 4y^2 = 1$. Then:

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