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Find the Centre, the Lengths of the Axes, Eccentricity, Foci of the Following Ellipse: 4x2 + 16y2 − 24x − 32y − 12 = 0 - Mathematics

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प्रश्न

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

4x2 + 16y2 − 24x − 32y − 12 = 0 

उत्तर

\[ 4 x^2 + 16 y^2 - 24x - 32y - 12 = 0\]
\[ \Rightarrow 4\left( x^2 - 6x \right) + 16\left( y^2 - 2y \right) = 12\]
\[ \Rightarrow 4\left( x^2 - 6x + 9 \right) + 16\left( y^2 - 2y + 1 \right) = 12 + 36 + 16\]
\[ \Rightarrow 4 \left( x - 3 \right)^2 + 16 \left( y - 1 \right)^2 = 64\]
\[ \Rightarrow \frac{\left( x - 3 \right)^2}{16} + \frac{\left( y - 1 \right)^2}{4} = 9\]
\[\text{ Centre }=\left( 3, 1 \right)\]
\[\text{ Major axis }=2a\]
\[ = 2 \times 4\]
\[ = 8\]
\[\text{ Minor axis }=2b\]
\[ = 2 \times 2\]
\[ = 4\]
\[e = \sqrt{1 - \frac{b^2}{a^2}}\]
\[ \Rightarrow e = \sqrt{1 - \frac{4}{16}}\]
\[ \Rightarrow e = \frac{\sqrt{3}}{2}\]
\[\text{ Foci } = \left( x \pm ae, y \right)\]
\[ = \left( 3 \pm 2\sqrt{3}, 1 \right)\]

 

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अध्याय 26: Ellipse - Exercise 26.1 [पृष्ठ २३]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 26 Ellipse
Exercise 26.1 | Q 10.5 | पृष्ठ २३

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