A) \[x-axis\]
B) \[y-axis\]
C) \[y+3=0\]
D) \[y-3=0\]
Correct Answer: B
Solution :
Key Idea The equation of the normal at the point\[({{x}_{1}},\,\,{{y}_{1}})\]to the ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]is \[\frac{x-{{x}_{1}}}{{{x}_{1}}/{{a}^{2}}}=\frac{y-{{y}_{1}}}{{{y}_{1}}/{{b}^{2}}}\] The equation of ellipse is \[9{{x}^{2}}+5{{y}^{2}}=45\] or \[\frac{{{x}^{2}}}{5}+\frac{{{y}^{2}}}{9}=1\] Here, \[{{a}^{2}}=5,\,\,{{b}^{2}}=9\] The equation of normal to the ellipse at the point \[(0,\,\,3)\]is \[\frac{x-0}{0/5}=\frac{y-3}{3/9}\] \[\Rightarrow \] \[x-0=0\] \[\Rightarrow \] \[x=0\] Which is the equation of \[y-\]axis. Alternative Given equation is \[9{{x}^{2}}+5{{y}^{2}}=45\] On differentiating, we get \[18x+10y\frac{dy}{dx}=0\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{18x}{10y}\] At\[(0,\,\,3),\,\,\left( \frac{dy}{dx} \right)=\frac{-18(0)}{10(3)}=0\] \[\therefore \] Equation of normal is \[y-3=-\frac{1}{0}(x-0)\] \[\Rightarrow \] \[x=0\] \[\Rightarrow \] \[y-axis\]
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