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If radii of director circles of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{(b)^{2}}=1$ are 2r and r respectively and e_e and e_h be the eccentricities of the ellipse and the hyperbola respectively then
Option: 1
$2e_{h}^{2}-e_{e}^{2}=6$

Option: 2
$e_{e}^{2}-4e_{h}^{2}=6$

Option: 3
$4e_{h}^{2}-e_{e}^{2}=6$

Option: 4
none of these

Answers (1)

best_answer

Eccentricity -

$e= \sqrt{1-\frac{b^{2}}{a^{2}}}$

- wherein

For the ellipse

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

Equation of director circles of ellipse and hyperbola are respectively.

$x^{2} + y^{2} = a^{2} + b^{2} \; and \; x^{2}+ y^{2} = a^{2}– b^{2}$

$a^{2} + b^{2} = 4r^{2} ......(i)$

$a^{2} + b^{2} = r^{2} ......(ii)$

$a^{2} =\frac{5r^{2}}{2},\;\; b^{2} =\frac{3r^{2}}{2}$

$e_{e}^{2}=1-\frac{b^{2}}{a^{2}}$

$\Rightarrow e_{e}^{2}=1-\frac{3r^{2}}{2}\times\frac{2}{5r^{2}}=1-\frac{3}{5}=\frac{2}{5}$

$e_{h}^{2}=1+\frac{b^{2}}{a^{2}}$

$e_{h}^{2}=1+\frac{3}{5}=\frac{8}{5}$

so $4e_{h}^{2}-e_{e}^{2}=4 \times\frac{8}{5}-\frac{2}{5}=\frac{30}{5}=6$

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Riya

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