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જો ઉપવલય $\frac{{{{\text{x}}^{\text{2}}}}}{{16}}\,\, + \;\,\frac{{{y^2}}}{{{b^2}}}\,\, = \,\,1\,\,\,$ ની નાભિઓ, અતિવલય $\frac{{{x^2}}}{{144}}\,\, - \,\,\frac{{{y^2}}}{{81}}\,\, = \,\,\frac{1}{{25}}$ ની નાભિઓને સમાન હોય,તો ${b^2}\, = \,\,...........$
$9$
$8$
$10$
$7$
Solution
Given hyperbola is, $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$
$\Rightarrow \frac{x^2}{144 / 25}-\frac{y^2}{81 / 25}=1$
$\Rightarrow a^2=\frac{144}{25}, b^2=\frac{81}{25}, e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{81}{144}}=\frac{15}{12}$
$\therefore$ foci of hyperbola are $(\pm ae , 0) i$ i. $(\pm 3,0)$
Now given ellipse is $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ $\Rightarrow a ^2=16$
Assume eccentricity of this ellipse is e' then its foci are $(\pm ae , 0) i . e \left(\pm 4 e ^{\prime}, 0\right)$ Given foci of given hyperbola and ellipse coincide
$\Rightarrow 4 e ^{\prime}=3 \Rightarrow e ^{\prime}=\frac{3}{4}$
For ellipse, using eccentricity relationship, $e ^2=1-\frac{ b ^2}{ a ^2}$
$\Rightarrow \frac{9}{16}=1-\frac{b^2}{16}$
$\therefore b^2=7$
