A) 5
B) 4
C) 3
D) 2
Correct Answer: D
Solution :
Since the ellipse and hyperbola intersect orthogonally then they are confocal. For ellipse \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4}=1,\] the foci are\[\left( \pm \,\sqrt{5},\,0 \right)\] and for hyperbola \[\frac{{{x}^{2}}}{9}\,+\frac{{{y}^{2}}}{4}=1,\] foci are \[\left( \pm \,\frac{2}{\alpha }\,\sqrt{1+{{\alpha }^{2}}}\,,\,\,0 \right)\] \[\therefore \,\,\frac{4}{{{\alpha }^{2}}}\,(1+{{\alpha }^{2}})=5\Rightarrow \,\frac{4}{{{\alpha }^{2}}}\,+4=5\]\[\Rightarrow \,{{\alpha }^{2}}=4\Rightarrow \,\alpha =2\]You need to login to perform this action.
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