The position of the point (1, 3) with respect | vedclass

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Question

(C) Given the equation of the ellipse: $4x^2 + 9y^2 - 16x - 54y + 61 = 0$.Rewrite the equation by completing the square:$4(x^2 - 4x) + 9(y^2 - 6y) + 6

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On the major axis

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(C) Given the equation of the ellipse: $4x^2 + 9y^2 - 16x - 54y + 61 = 0$.Rewrite the equation by completing the square:$4(x^2 - 4x) + 9(y^2 - 6y) + 61 = 0$$4(x^2 - 4x + 4) + 9(y^2 - 6y + 9) = -61 + 16 + 81$$4(x - 2)^2 + 9(y - 3)^2 = 36$Divide by $36$:$\frac{(x - 2)^2}{9} + \frac{(y - 3)^2}{4} = 1$.This is an ellipse with center $(2, 3)$ and major axis along the line $y = 3$.Substitute the point $(1, 3)$ into the equation $f(x, y) = 4x^2 + 9y^2 - 16x - 54y + 61$:$f(1, 3) = 4(1)^2 + 9(3)^2 - 16(1) - 54(3) + 61$$f(1, 3) = 4 + 81 - 16 - 162 + 61 = -32$.Since $f(1, 3) However,checking the major axis $y = 3$,the point $(1, 3)$ satisfies $y = 3$,so it lies on the major axis.

The position of the point $(1, 3)$ with respect to the ellipse $4x^2 + 9y^2 - 16x - 54y + 61 = 0$ is:

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