If x^2 = 8ay is the transformed equation of x | vedclass

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(C) The given equation is $x^2 + 6x - 4y + 15 = 0$. Completing the square for $x$: $(x^2 + 6x + 9) - 9 - 4y + 15 = 0$. $(x + 3)^2 - 4y + 6 = 0$. $(x +

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$12$

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(C) The given equation is $x^2 + 6x - 4y + 15 = 0$. Completing the square for $x$: $(x^2 + 6x + 9) - 9 - 4y + 15 = 0$. $(x + 3)^2 - 4y + 6 = 0$. $(x + 3)^2 = 4y - 6$. $(x + 3)^2 = 4(y - \frac{3}{2})$. Comparing this with the transformed equation $(x - \alpha)^2 = 8a(y - \beta)$,we have: $x - \alpha = x + 3 \Rightarrow \alpha = -3$. $y - \beta = y - \frac{3}{2} \Rightarrow \beta = \frac{3}{2}$. $8a = 4 \Rightarrow a = \frac{1}{2}$. Now,calculate $2\alpha + 8\beta^2$: $2(-3) + 8(\frac{3}{2})^2 = -6 + 8(\frac{9}{4}) = -6 + 18 = 12$.

If $x^2 = 8ay$ is the transformed equation of $x^2 - 4y + 6x + 15 = 0$ when the origin is shifted to the point $(\alpha, \beta)$ by translation of axes,then $2\alpha + 8\beta^2 =$

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