A) Eccentricity
B) Abscissa of foci
C) Directrix
D) Vertex
Correct Answer: B
Solution :
\[{{e}^{2}}=1+\frac{{{b}^{2}}}{a2}=1+\frac{{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha }=\frac{1}{{{\cos }^{2}}\alpha }\] and \[{{a}^{2}}={{\cos }^{2}}\alpha \] \[\therefore \]\[{{a}^{2}}{{e}^{2}}=1\] \[\therefore \]\[foci(\pm ae,\,\,0)=(\pm 1,\,\,0)\], which Is independent of\[\alpha \].You need to login to perform this action.
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