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Question

(B) Let the origin be shifted to $(h, k)$. The transformation equations are $x = x' + h$ and $y = y' + k$. Substituting these into the given equation

Vedclass · Accepted Answer

$h=-\frac{10}{17}, k=-\frac{13}{17}$

Vedclass · Answer

(B) Let the origin be shifted to $(h, k)$. The transformation equations are $x = x' + h$ and $y = y' + k$. Substituting these into the given equation $x^2 + 5xy + 2y^2 + 5x + 6y + 7 = 0$ gives: $(x' + h)^2 + 5(x' + h)(y' + k) + 2(y' + k)^2 + 5(x' + h) + 6(y' + k) + 7 = 0$. Expanding this,the terms of the first degree in $x'$ and $y'$ are: $(2h + 5k + 5)x' + (5h + 4k + 6)y'$. For the equation to be free from first-order terms,these coefficients must be zero: $2h + 5k + 5 = 0$ $(i)$ $5h + 4k + 6 = 0$ (ii) Multiplying $(i)$ by $4$ and (ii) by $5$: $8h + 20k + 20 = 0$ $25h + 20k + 30 = 0$ Subtracting the first from the second: $17h + 10 = 0 \implies h = -\frac{10}{17}$. Substituting $h$ into $(i)$: $2(-\frac{10}{17}) + 5k + 5 = 0 \implies -\frac{20}{17} + 5k + 5 = 0 \implies 5k = \frac{20}{17} - 5 = \frac{20-85}{17} = -\frac{65}{17} \implies k = -\frac{13}{17}$. Thus,$h = -\frac{10}{17}$ and $k = -\frac{13}{17}$.

If the origin is shifted to a point $(h, k)$ by translation of axes in order to make the equation $x^2+5xy+2y^2+5x+6y+7=0$ free from first-order terms,then:

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