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Find the Centre, the Lengths of the Axes, Eccentricity, Foci of the Following Ellipse: X2 + 2y2 − 2x + 12y + 10 = 0 - Mathematics

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प्रश्न

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

x2 + 2y2 − 2x + 12y + 10 = 0 

उत्तर

\[ x^2 + 2 y^2 - 2x + 12y + 10 = 0\]
\[ \Rightarrow \left( x^2 - 2x \right) + 2\left( y^2 + 6y \right) = - 10\]
\[ \Rightarrow \left( x^2 - 2x + 1 \right) + 2\left( y^2 + 6y + 9 \right) = - 10 + 18 + 1\]
\[ \Rightarrow \left( x - 1 \right)^2 + 2 \left( y + 3 \right)^2 = 9\]
\[ \Rightarrow \frac{\left( x - 1 \right)^2}{9} + \frac{\left( y + 3 \right)^2}{\frac{9}{2}} = 9\]
\[\text{ Here }, x_1 = 1 \text{ and } y_1 = - 3\]
\[\text{ Also }, a = 3 \text{ and } b = \frac{3}{\sqrt{2}}\]
\[\text{ Centre }=\left( 1, - 3 \right)\]
\[\text{ Major axis }=2a\]
\[ \Rightarrow 2 \times 3 = 6\]
\[\text{ Minor axis }=2b\]
\[ \Rightarrow 2 \times \frac{3}{\sqrt{2}} = 3\sqrt{2}\]
\[e = \sqrt{1 - \frac{b^2}{a^2}}\]
\[ \Rightarrow e = \sqrt{1 - \frac{\frac{9}{2}}{9}}\]
\[ \Rightarrow e = \frac{1}{\sqrt{2}}\]
\[\text{ Foci } = \left( x_1 \pm ae, y_1 \right)\]
\[ = \left( 1 \pm \frac{3}{\sqrt{2}}, - 3 \right)\]

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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 10.1 | पृष्ठ २३

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