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

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(C) To determine the position of a point $(x_1, y_1)$ with respect to the ellipse $S(x, y) = 4x^2 + 5y^2 - 1 = 0$,we evaluate the expression $S(x_1, y

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Is outside the curve

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(C) To determine the position of a point $(x_1, y_1)$ with respect to the ellipse $S(x, y) = 4x^2 + 5y^2 - 1 = 0$,we evaluate the expression $S(x_1, y_1)$:$(i)$ If $S(x_1, y_1) = 0$,the point lies on the ellipse.$(ii)$ If $S(x_1, y_1) > 0$,the point lies outside the ellipse.$(iii)$ If $S(x_1, y_1) Given the point $(4, -3)$ and the ellipse equation $4x^2 + 5y^2 - 1 = 0$,we substitute the coordinates:$S(4, -3) = 4(4)^2 + 5(-3)^2 - 1$$S(4, -3) = 4(16) + 5(9) - 1$$S(4, -3) = 64 + 45 - 1$$S(4, -3) = 108$Since $108 > 0$,the point $(4, -3)$ lies outside the ellipse.

The point $(4, -3)$ with respect to the ellipse $4x^2 + 5y^2 = 1$ is:

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