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Eccentricity of Ellipse

E 1: x 2/ a 2...

Question

$E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0, (a > b)$ and $E_{2} : \frac{x^{2}}{k^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0, (k < b)$ is inscribed in $E_{1} .$ If $E_{1}$ and $E_{2}$ have same eccentricities and length of minor axis of $E_{2} = p \times L L R of E_{1},$ then $p =$

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Solution

$E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0, (a > b)$ and $E_{2} : \frac{x^{2}}{k^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0, (k < b)$

Let $e_{1}$ and $e_{2}$ are the eccentricities of ellipse $E_{1}$ and $E_{2},$ then given that,
$e_{1} = e_{2} \Rightarrow e_{1}^{2} = e_{2}^{2} \Rightarrow 1 - \frac{b^{2}}{a^{2}} = 1 - \frac{k^{2}}{b^{2}}$
$\Rightarrow k = \frac{b^{2}}{a} \Rightarrow 2 k = \frac{2 b^{2}}{a} \Rightarrow length of minor axis of E_{2} = 1 \times [LLR of E_{1}] ∴ p = 1$

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Similar questions

E_{1} = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0, (a > b)

E_{2} = \frac{x^{2}}{k^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0, (k < b) E_{2}

is inscribed in

E_{1}

E_{1}

E_{2}

have same eccentricities then length of minor axis of

E_{2} = p (

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p = ?

E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b .

E_{2}

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E_{1}

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\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1

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\frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}} = 1

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Eccentricity of Ellipse

Standard XII Mathematics

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