A) \[y-3=0\]
B) \[y+3=0\]
C) x-axis
D) y-axis
Correct Answer: D
Solution :
For \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1,\] equation of normal at point\[({{x}_{1}},{{y}_{1}})\], Þ\[\frac{(x-{{x}_{1}}){{a}^{2}}}{{{x}_{1}}}=\frac{(y-{{y}_{1}}){{b}^{2}}}{{{y}_{1}}}\]; \\[({{x}_{1}},{{y}_{1}})\equiv (0,3),\,\,{{a}^{2}}=5,\,\,{{b}^{2}}=9\] \[\Rightarrow \frac{(x-0)}{0}\,\,\,5=\frac{(y-3).9}{3}\] or \[x=0\] i.e., y-axis.
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