A) 3
B) 4
C) 1/2
D) 2
Correct Answer: D
Solution :
\[\left( \pm \,ae,\frac{{{b}^{2}}}{a} \right)\]are extremities of the latus-rectum having positive ordinates which lies on parabola\[{{x}^{2}}=-2(y-2).\] \[\Rightarrow \] \[{{a}^{2}}{{e}^{2}}=-2\left( \frac{{{b}^{2}}}{a}-2 \right)\] ?(i) Also \[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\] From (i) and (i), we get \[{{a}^{2}}{{e}^{2}}-2a{{e}^{2}}+2a-4=0\] \[\Rightarrow \]\[a{{e}^{2}}(a-2)+2(a-2)=0\] \[\therefore \]\[(a{{e}^{2}}+2)(a-2)=0\] Hence, \[a=2.\]You need to login to perform this action.
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