The point (1,3) with respect to the ellipse 4 | vedclass

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(D) The given equation of the ellipse is $4x^2+9y^2-16x-54y+61=0$.Rewriting the equation by completing the square:$4(x^2-4x)+9(y^2-6y)=-61$$4(x^2-4x+4

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on the major axis

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(D) The given equation of the ellipse is $4x^2+9y^2-16x-54y+61=0$.Rewriting the equation by completing the square:$4(x^2-4x)+9(y^2-6y)=-61$$4(x^2-4x+4)+9(y^2-6y+9)=-61+16+81$$4(x-2)^2+9(y-3)^2=36$Dividing by $36$,we get:$\frac{(x-2)^2}{9}+\frac{(y-3)^2}{4}=1$This is an ellipse with center $(2,3)$ and major axis along the line $y=3$.The point $(1,3)$ satisfies the equation $y=3$,which is the equation of the major axis.Since the $x$-coordinate $1$ is between the vertices $(-1,3)$ and $(5,3)$,the point $(1,3)$ lies inside the ellipse on the major axis.

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

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