Let (x, y) be a variable point on the curve 4 | vedclass

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(A) Given the equation of the ellipse: $4x^2 + 9y^2 - 8x - 36y + 15 = 0$. Completing the square,we get: $4(x^2 - 2x + 1) + 9(y^2 - 4y + 4) = -15 + 4 +

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$\frac{325}{36}$

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(A) Given the equation of the ellipse: $4x^2 + 9y^2 - 8x - 36y + 15 = 0$. Completing the square,we get: $4(x^2 - 2x + 1) + 9(y^2 - 4y + 4) = -15 + 4 + 36$. $4(x - 1)^2 + 9(y - 2)^2 = 25$. Dividing by $25$,we have: $\frac{(x - 1)^2}{25/4} + \frac{(y - 2)^2}{25/9} = 1$. Let $X = x - 1$ and $Y = y - 2$. The expression becomes $X^2 + Y^2$. This represents the square of the distance from the center $(1, 2)$ to any point $(x, y)$ on the ellipse. The distance squared $r^2 = X^2 + Y^2$ ranges from the square of the semi-minor axis to the square of the semi-major axis. Here,$a^2 = \frac{25}{4} = 6.25$ and $b^2 = \frac{25}{9} \approx 2.77$. Thus,$\min(X^2 + Y^2) = b^2 = \frac{25}{9}$ and $\max(X^2 + Y^2) = a^2 = \frac{25}{4}$. The sum is $\frac{25}{9} + \frac{25}{4} = 25 \left( \frac{4 + 9}{36} ight) = \frac{25 	imes 13}{36} = \frac{325}{36}$.

Let $(x, y)$ be a variable point on the curve $4x^2 + 9y^2 - 8x - 36y + 15 = 0$. Then,$\min (x^2 - 2x + y^2 - 4y + 5) + \max (x^2 - 2x + y^2 - 4y + 5)$ is

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