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Radius of Gyration

Let the major...

Question

Let the major axis of a standard ellipse equals the transverse axis of a standard hyperbola director circles have radius equal to 2R and R respectively. If $e_{1}$ and $e_{2}$ , are the eccentrieities of the ellipse and hyperbola then the correct relation is

A

4 e_{1}^{2} - e_{2}^{2} = 6

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B

e_{1}^{2} - 4 e_{2}^{2} = 2

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C

4 e_{1}^{2} - e_{1}^{2} = 6

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D

2 e_{1}^{2} - e_{2}^{2} = 4

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Solution

The correct option is A $4 e_{1}^{2} - e_{2}^{2} = 6$
$L e t A a n d B b e t h e m i n o r a n d m a j o r a x i s o f e l l i p s e A n d, a a n d b b e t t h e t r a n s v e r s e a n d c o n j u g a t e a x i s o f h y p e r b o l a . G i v e n : B = a A^{2} + B^{2} = 4 R^{2} a^{2} - b^{2} = R^{2} O r, A^{2} + B^{2} = 4 (a^{2} - b^{2}) O n d i v i d i n g b y B a n d u s i n g B = a w e g e t : 1 + \frac{A^{2}}{B^{2}} = 4 (1 - \frac{b^{2}}{a^{2}}) - - - - - - (1) W e k n o w, {(e 1)}^{2} = 1 - \frac{A^{2}}{B^{2}} O r, \frac{A^{2}}{B^{2}} = 1 - {(e 1)}^{2} - - - - - - (2) {(e 2)}^{2} = 1 + \frac{b^{2}}{a^{2}} O r, \frac{b^{2}}{a^{2}} = {(e 2)}^{2} - 1 - - - - - - - (3) U s i n g e q u a t i o n s 123 w e g e t : 1 + 1 - {(e 1)}^{2} = 4 (1 - ({(e 2)}^{2} - 1)) O r, 2 - {(e 1)}^{2} = 8 - 4 {(e 2)}^{2} O r, 4 {(e 2)}^{2} - {(e 1)}^{2} = 6$

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

\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1

are having a same foci and length of minor axis of ellipse is same as the conjugate axis of the hyperbola. If

e_{1}

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e_{1}

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

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

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

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