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Find the Equation of the Ellipse in the Case: (I) Eccentricity E = 1 2 and Foci (± 2, 0) - Mathematics

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प्रश्न

Find the equation of the ellipse in the case:

eccentricity e = \[\frac{1}{2}\] and foci (± 2, 0)

संक्षेप में उत्तर
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उत्तर

\[e = \frac{1}{2} \text{ and foci }  = ( \pm 2, 0)\]
\[\text{ Coordinates of the foci } = \left( \pm ae, 0 \right)\]
\[\text{ We have ae } = 2 \]
\[ \Rightarrow a \times \frac{1}{2} = 2\]
\[ \Rightarrow a = 4\]
\[\text{ Now, }  e = \sqrt{1 - \frac{b^2}{a^2}}\]
\[ \Rightarrow \frac{1}{2} = \sqrt{1 - \frac{b^2}{16}}\]
\[\text{ On squaring both sides, we get:} \]
\[\frac{1}{4} = \frac{16 - b^2}{16}\]
\[ \Rightarrow 64 - 4 b^2 = 16\]
\[ \Rightarrow b^2 = \frac{48}{4}\]
\[ \Rightarrow b^2 = 12\]
\[\text{ Now } , \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]
\[ \Rightarrow \frac{x^2}{16} + \frac{y^2}{12} = 1\]
\[ \Rightarrow \frac{3 x^2 + 4 y^2}{48} = 1\]
\[ \Rightarrow 3 x^2 + 4 y^2 = 48\]
\[\text{ This is the required equation of the ellipse.} \]

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अध्याय 26: Ellipse - Exercise 26.1 [पृष्ठ २२]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 26 Ellipse
Exercise 26.1 | Q 5.01 | पृष्ठ २२
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