The centre of the ellipse 4x^2 + 9y^2 - 16x - | vedclass

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(B) Given equation: $4x^2 + 9y^2 - 16x - 54y + 61 = 0$Rearranging the terms: $(4x^2 - 16x) + (9y^2 - 54y) = -61$Factor out the coefficients: $4(x^2 -

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$(2, 3)$

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(B) Given equation: $4x^2 + 9y^2 - 16x - 54y + 61 = 0$Rearranging the terms: $(4x^2 - 16x) + (9y^2 - 54y) = -61$Factor out the coefficients: $4(x^2 - 4x) + 9(y^2 - 6y) = -61$Complete the square: $4(x^2 - 4x + 4) + 9(y^2 - 6y + 9) = -61 + 16 + 81$Simplify: $4(x - 2)^2 + 9(y - 3)^2 = 36$Divide by $36$: $\frac{(x - 2)^2}{9} + \frac{(y - 3)^2}{4} = 1$Comparing this with the standard form $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$,the centre $(h, k)$ is $(2, 3).$

The centre of the ellipse $4x^2 + 9y^2 - 16x - 54y + 61 = 0$ is

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