A) Eccentricity
B) Directrix
C) Abscissae of vertices
D) Abscissae of foci
Correct Answer: D
Solution :
Use the eccentricity formula i.e., \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}\] The given equation of hyperbola is \[\frac{{{x}^{2}}}{{{\cos }^{2}}\alpha }-\frac{{{y}^{2}}}{{{\sin }^{2}}\alpha }=1\] Here,\[{{a}^{2}}=co{{s}^{2}}\alpha \]and \[{{b}^{2}}=si{{n}^{2}}\alpha \] Coordinates of foci are \[(\pm \,ae,\,\,0)\]. \[\because \] \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}\] \[\Rightarrow \] \[e=\sqrt{1+\frac{{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha }}=\sqrt{1+{{\tan }^{2}}\alpha }\] \[\Rightarrow \] \[e=\sec \alpha \] Hence, abscissae of foci remains constant when \[\alpha \] varies.
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