A) - 7
B) - 5
C) 5
D) 7
Correct Answer: A
Solution :
\[9{{e}^{2}}-18e+5=0\]\[\Rightarrow \]\[e=\frac{5}{3}\] \[\therefore \]\[1+\frac{{{b}^{2}}}{{{a}^{2}}}={{e}^{2}}=\frac{25}{9}\] ?(i) Also distance between foci and directrix is \[=\left( ae-\frac{a}{e} \right)=5-\frac{9}{5}\] \[\Rightarrow \]\[a\left( \frac{5}{3}-\frac{3}{5} \right)=\frac{16}{5}\Rightarrow a=3\]from (i) \[1+\frac{{{b}^{2}}}{9}={{e}^{2}}=\frac{25}{9}\Rightarrow {{b}^{2}}=16\] \[\therefore \]\[{{a}^{2}}-{{b}^{2}}=9-16=-7\]You need to login to perform this action.
You will be redirected in
3 sec