Find the centre, the lengths of the axes, eccentricity, foci of the following ellipse:

4*x*^{2} + 16*y*^{2} − 24*x* − 32*y* − 12 = 0

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#### Solution

\[ 4 x^2 + 16 y^2 - 24x - 32y - 12 = 0\]

\[ \Rightarrow 4\left( x^2 - 6x \right) + 16\left( y^2 - 2y \right) = 12\]

\[ \Rightarrow 4\left( x^2 - 6x + 9 \right) + 16\left( y^2 - 2y + 1 \right) = 12 + 36 + 16\]

\[ \Rightarrow 4 \left( x - 3 \right)^2 + 16 \left( y - 1 \right)^2 = 64\]

\[ \Rightarrow \frac{\left( x - 3 \right)^2}{16} + \frac{\left( y - 1 \right)^2}{4} = 9\]

\[\text{ Centre }=\left( 3, 1 \right)\]

\[\text{ Major axis }=2a\]

\[ = 2 \times 4\]

\[ = 8\]

\[\text{ Minor axis }=2b\]

\[ = 2 \times 2\]

\[ = 4\]

\[e = \sqrt{1 - \frac{b^2}{a^2}}\]

\[ \Rightarrow e = \sqrt{1 - \frac{4}{16}}\]

\[ \Rightarrow e = \frac{\sqrt{3}}{2}\]

\[\text{ Foci } = \left( x \pm ae, y \right)\]

\[ = \left( 3 \pm 2\sqrt{3}, 1 \right)\]

\[ \Rightarrow 4\left( x^2 - 6x \right) + 16\left( y^2 - 2y \right) = 12\]

\[ \Rightarrow 4\left( x^2 - 6x + 9 \right) + 16\left( y^2 - 2y + 1 \right) = 12 + 36 + 16\]

\[ \Rightarrow 4 \left( x - 3 \right)^2 + 16 \left( y - 1 \right)^2 = 64\]

\[ \Rightarrow \frac{\left( x - 3 \right)^2}{16} + \frac{\left( y - 1 \right)^2}{4} = 9\]

\[\text{ Centre }=\left( 3, 1 \right)\]

\[\text{ Major axis }=2a\]

\[ = 2 \times 4\]

\[ = 8\]

\[\text{ Minor axis }=2b\]

\[ = 2 \times 2\]

\[ = 4\]

\[e = \sqrt{1 - \frac{b^2}{a^2}}\]

\[ \Rightarrow e = \sqrt{1 - \frac{4}{16}}\]

\[ \Rightarrow e = \frac{\sqrt{3}}{2}\]

\[\text{ Foci } = \left( x \pm ae, y \right)\]

\[ = \left( 3 \pm 2\sqrt{3}, 1 \right)\]

Is there an error in this question or solution?

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