The point to which the origin is to be shifte | vedclass

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(A) Given equation is $y^2+4y+8x-2=0$. Let the origin be shifted to $(\alpha, \beta)$. Then,$x = X + \alpha$ and $y = Y + \beta$. Substituting these i

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$\left(\frac{3}{4}, -2\right)$

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(A) Given equation is $y^2+4y+8x-2=0$. Let the origin be shifted to $(\alpha, \beta)$. Then,$x = X + \alpha$ and $y = Y + \beta$. Substituting these into the original equation: $(Y + \beta)^2 + 4(Y + \beta) + 8(X + \alpha) - 2 = 0$ $Y^2 + 2Y\beta + \beta^2 + 4Y + 4\beta + 8X + 8\alpha - 2 = 0$ $Y^2 + Y(2\beta + 4) + 8X + (\beta^2 + 4\beta + 8\alpha - 2) = 0$. For the transformed equation to have no $Y$ term and no constant term,we set their coefficients to zero: $2\beta + 4 = 0 \Rightarrow \beta = -2$. $\beta^2 + 4\beta + 8\alpha - 2 = 0$. Substituting $\beta = -2$: $(-2)^2 + 4(-2) + 8\alpha - 2 = 0$ $4 - 8 + 8\alpha - 2 = 0$ $8\alpha - 6 = 0 \Rightarrow \alpha = \frac{6}{8} = \frac{3}{4}$. Thus,the origin is shifted to $\left(\frac{3}{4}, -2\right)$.

The point to which the origin is to be shifted by translation of axes so that the transformed equation of $y^2+4y+8x-2=0$ will not contain $y$ term and constant term is

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