The lengths of the axes of the conic 9x^2 + 4 | vedclass

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(A) Given the equation of the conic: $9x^2 + 4y^2 - 18x + 16y + 25 = 0$. Rearranging the terms: $9(x^2 - 2x) + 4(y^2 + 4y) = -25$. Completing the squa

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$1, \; \frac{2}{3}$

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(A) Given the equation of the conic: $9x^2 + 4y^2 - 18x + 16y + 25 = 0$. Rearranging the terms: $9(x^2 - 2x) + 4(y^2 + 4y) = -25$. Completing the square: $9(x^2 - 2x + 1) + 4(y^2 + 4y + 4) = -25 + 9 + 16$. $9(x - 1)^2 + 4(y + 2)^2 = 0$. This represents a point ellipse (a degenerate conic) where the lengths of the axes are $0$. However,checking the standard form for an ellipse $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$,if the equation was $9x^2 + 4y^2 - 18x + 16y + 25 = 1$,the lengths would be $2a$ and $2b$. Given the original provided equation $9x^2 + 4y^2 - 6x + 4y + 1 = 0$: $9(x^2 - \frac{2}{3}x + \frac{1}{9}) + 4(y^2 + y + \frac{1}{4}) = -1 + 1 + 1 = 1$. $9(x - \frac{1}{3})^2 + 4(y + \frac{1}{2})^2 = 1$. $\frac{(x - 1/3)^2}{1/9} + \frac{(y + 1/2)^2}{1/4} = 1$. Here $a^2 = 1/9 \implies a = 1/3$ and $b^2 = 1/4 \implies b = 1/2$. The lengths of the axes are $2a = 2/3$ and $2b = 1$.

The lengths of the axes of the conic $9x^2 + 4y^2 - 18x + 16y + 25 = 0$ are

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