A) 9
B) 144
C) 12
D) 4
E) 25
Correct Answer: C
Solution :
The equation of ellipse is \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] ?. (i) and equation of hyperbola is \[\frac{{{x}^{2}}}{100}-\frac{4{{y}^{2}}}{225}=1\] ...(ii) For ellipse, \[e=\sqrt{1-\frac{{{b}^{2}}}{16}}\] and equation of directrices are \[x=\pm \left( \frac{4}{\sqrt{1-\frac{{{b}^{2}}}{16}}} \right)\] \[\left( \because x=\pm \frac{a}{e} \right)\] and for hyperbola, \[e=\sqrt{1+\frac{225}{400}}\] \[=\sqrt{\frac{625}{400}}=\frac{25}{20}\] \[\therefore \]Equation of directrices area\[=\pm \left( \frac{10\times 20}{25} \right)\] \[=\pm 8\] Given that, ellipse and hyperbola have same directrices: \[\therefore \] \[\frac{4}{\sqrt{1-\frac{{{b}^{2}}}{16}}}=8\Rightarrow \frac{1}{4}=1-\frac{{{b}^{2}}}{16}\] \[\Rightarrow \] \[\frac{{{b}^{2}}}{16}=1-\frac{1}{4}=\frac{3}{4}\] \[\Rightarrow \] \[{{b}^{2}}=3\times 4=12\]
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