If the equation of a conic section is $\frac{{(x - 3)}^{2}}{49} + \frac{{(y - 4)}^{2}}{13} = 1$ $(x-3)^2/49+(y-4)^2/13=1$ , how has its center been translated?

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If the equation of a conic section is $\frac{{(x - 3)}^{2}}{49} + \frac{{(y - 4)}^{2}}{13} = 1$ $(x-3)^2/49+(y-4)^2/13=1$ , how has its center been translated?

1 Answer

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Answer 1 · 2015-09-23T18:13:17

This conic section is called an ellipse with a center at (3, 4).

Explanation:

This is an ellipse with a horizontal major axis centered at (3, 4). This ellipse has both a horizontal and vertical shift from a standard ellipse centered at (0,0).

It has been translated 3 units horizontally to the right and 4 units vertically upward .

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If the equation of a conic section is $\frac{{(x - 3)}^{2}}{49} + \frac{{(y - 4)}^{2}}{13} = 1$ $(x-3)^2/49+(y-4)^2/13=1$ , how has its center been translated?

1 Answer

Explanation:

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If the equation of a conic section is (x−3)249+(y−4)213=1(x-3)^2/49+(y-4)^2/13=1, how has its center been translated?

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Explanation:

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If the equation of a conic section is $\frac{{(x - 3)}^{2}}{49} + \frac{{(y - 4)}^{2}}{13} = 1$ $(x-3)^2/49+(y-4)^2/13=1$ , how has its center been translated?