A) \[-(2am+b{{m}^{2}})\]
B) \[\frac{({{a}^{2}}+{{b}^{2}})m}{\sqrt{{{a}^{2}}+{{b}^{2}}{{m}^{2}}}}\]
C) \[-\frac{({{a}^{2}}-{{b}^{2}})m}{\sqrt{{{a}^{2}}+{{b}^{2}}{{m}^{2}}}}\]
D) \[\frac{({{a}^{2}}-{{b}^{2}})m}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
Correct Answer: C
Solution :
As we know that the line \[lx+my+n=0\] is normal to \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\], if \[\frac{{{a}^{2}}}{{{l}^{2}}}+\frac{{{b}^{2}}}{{{m}^{2}}}=\frac{{{({{a}^{2}}-{{b}^{2}})}^{2}}}{{{n}^{2}}}\]. But in this condition, we have to replace l by m, m by ?1 and n by c, then the required condition is \[c=\pm \frac{({{a}^{2}}-{{b}^{2}})m}{\sqrt{{{a}^{2}}+{{b}^{2}}{{m}^{2}}}}\].
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