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Find the center, vertices, foci, and eccentricity of the ellipse. \[ 16 x^{2}+24 y^{2}-16 x+48 y-68=0 \] center \( (x, y)=( \) \( \square \) ) vertices \( \quad(x, y)=( \) \( \square \) )(smaller \( x \)-value) \[ (x, y)=(\square) \text { (larger } x \text {-value) } \] foci \( (x, y)= \) \( \square \) ) (smaller \( x \)-value) \[ \begin{array}{l} (x, y)=(\square \\ \square \end{array} \] eccentricity

Ask by Collins Ball. in the United States
Mar 14,2025

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Center: \( \left( \frac{1}{2}, -1 \right) \) Vertices: - Smaller \( x \)-value: \( \left( \frac{1}{2} - \sqrt{6}, -1 \right) \) - Larger \( x \)-value: \( \left( \frac{1}{2} + \sqrt{6}, -1 \right) \) Foci: - Smaller \( x \)-value: \( \left( \frac{1}{2} - \sqrt{2}, -1 \right) \) - Larger \( x \)-value: \( \left( \frac{1}{2} + \sqrt{2}, -1 \right) \) Eccentricity: \( \frac{\sqrt{3}}{3} \)

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The Deep Dive

To find the center, vertices, foci, and eccentricity of the ellipse given by the equation \(16 x^{2}+24 y^{2}-16 x+48 y-68=0\), we first rewrite the equation in its standard form. 1. Rearranging the terms: \[ 16x^{2} - 16x + 24y^{2} + 48y - 68 = 0 \] Dividing the whole equation by 4: \[ 4x^{2} - 4x + 6y^{2} + 12y - 17 = 0 \] 2. Completing the square for \(x\) and \(y\): - For \(x\): \[ 4(x^{2} - x) = 4\left(x - \frac{1}{2}\right)^{2} - 4\left(\frac{1}{2}\right)^{2} = 4\left(x - \frac{1}{2}\right)^{2} - 1 \] - For \(y\): \[ 6(y^{2} + 2y) = 6\left(y + 1\right)^{2} - 6 \] Plugging the squares back into our equation: \[ 4\left(x - \frac{1}{2}\right)^{2} - 1 + 6\left(y + 1\right)^{2} - 6 - 17 = 0 \] \[ 4\left(x - \frac{1}{2}\right)^{2} + 6\left(y + 1\right)^{2} - 24 = 0 \] \[ 4\left(x - \frac{1}{2}\right)^{2} + 6\left(y + 1\right)^{2} = 24 \] Dividing by 24 to get it into standard form: \[ \frac{\left(x - \frac{1}{2}\right)^{2}}{6} + \frac{\left(y + 1\right)^{2}}{4} = 1 \] 3. From the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\): - The center \((h, k) = \left(\frac{1}{2}, -1\right)\). - The semi-major axis \(a = \sqrt{6}\) (along the x-axis) and semi-minor axis \(b = \sqrt{4} = 2\). 4. Finding the vertices: \[ \text{Vertices} = \left(\frac{1}{2} - \sqrt{6}, -1\right) \text{ and } \left(\frac{1}{2} + \sqrt{6}, -1\right) \] - Smaller \(x\)-value: \(\left(\frac{1}{2} - \sqrt{6}, -1\right)\) - Larger \(x\)-value: \(\left(\frac{1}{2} + \sqrt{6}, -1\right)\) 5. Finding the foci: - The distance \(c = \sqrt{a^2 - b^2} = \sqrt{6 - 4} = \sqrt{2}\). \[ \text{Foci} = \left(\frac{1}{2} - \sqrt{2}, -1\right) \text{ and } \left(\frac{1}{2} + \sqrt{2}, -1\right) \] - Smaller \(x\)-value: \(\left(\frac{1}{2} - \sqrt{2}, -1\right)\) - Larger \(x\)-value: \(\left(\frac{1}{2} + \sqrt{2}, -1\right)\) 6. Eccentricity \(e = \frac{c}{a} = \frac{\sqrt{2}}{\sqrt{6}} = \frac{\sqrt{12}}{6} = \frac{1}{3}\). Now, presenting the final results: center \( (x, y)=( \frac{1}{2} , -1) \) vertices \( \quad(x, y)=( \frac{1}{2} - \sqrt{6} , -1) ) \text{ (

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