Question

Graphing Conic SectionsExercises 63-68 give equations for conic sections and tell how many units up or down and to the right or left each curve is to be shifted. Find an equation for the new conic section, and find the new foci, vertices, centers, and asymptotes, as appropriate. If the curve is a parabola, find the new directrix as well.x²/169 + y²/144 = 1, right 5, up 12

Accepted Answer

Identify the type of conic section given by the equation $$\frac{x$$^{2}$$}{169} + \frac{y$$^{2}$$}{144} = 1. $$Since both $$x^{2}$$ and $$y^{2}$$ terms are positive and the equation equals 1, this is an ellipse centered at the origin $$(0,0). $$Recall the standard form of an ellipse centered at $$(h,k)$$: $$\frac{(x - h$$)^{2}$$}{a$$^{2}$$} + \frac{(y - k$$)^{2}$$}{b$$^{2}$$} = 1. $$Here, $$a^{2} = 169$ and b$$^{2}$$ = 144$$, so $$a = 13$$ and $$b = 12. $$Apply the given shifts: right 5 units and up 12 units. This changes the center from $$(0,0) to$$ $$(5,12). $$Substitute $$h=5$$ and $$k=12$$ into the ellipse equation to get the new equation: $$\frac{(x - 5$$)^{2}$$}{169} + \frac{(y - 12$$)^{2}$$}{144} = 1. $$Find the new foci. For an ellipse, the focal distance $$c is $$given by $$c = \sqrt{a$$^{2}$$ - b$$^{2}$$}. $$Calculate $$c$$ and then determine the coordinates of the foci relative to the new center $$(5,12). $$Since $$a^{2}$$ \u003e $$b^{2}$$, the major axis is along the x-axis, so the foci are at $$(h \pm c, k). $$Find the vertices. The vertices lie along the major axis at a distance $$a$$ from the center. So, the vertices are at $$(h \pm a, k). $$Since this is an ellipse, there are no asymptotes or directrix to find.

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