A) all \[m\in R\]
B) all me [-1, 1]
C) all \[m\in R-\{0\}\]
D) None of these
Correct Answer: C
Solution :
\[{{y}^{2}}=4a\left( \frac{y+am}{m} \right)\]i.e., \[m{{y}^{2}}-4ay-4{{a}^{2}}m=0\] \[m\ne 0;16{{a}^{2}}+16{{a}^{2}}{{m}^{2}}>0\]which is true \[\forall \]m. \[\therefore \]\[m\in R-\{0\}\]
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